AP-Level Physics 2

511 metres per second, in the air you are breathing

At 20 °C, a typical nitrogen molecule in the air around you is moving at 511 m/s, about one and a half times the speed of sound in that same air. It does not keep going for long. It hits another molecule and changes direction, then hits another, and the molecules next to a wall hit the wall. Each impact on the wall is a tiny push. Add up the pushes on one square metre of wall during one second and you have the number a barometer reads.

This lesson starts from one atom bouncing off one wall, uses nothing but the impulse-momentum theorem from your first physics course, and arrives at two facts that the whole of Unit 9 rests on. Pressure is the average force of molecular impacts per unit area. Temperature measures the average kinetic energy of the molecules. The ideal gas law, the first law of thermodynamics and the second law are all built on those two sentences.

What a pressure is, and how big one atmosphere is

A gas pushes on everything it touches, and the natural way to describe that push is per unit area. Pressure is the perpendicular force on a surface divided by the area of the surface:

P = F⊥/A

The unit is the newton per square metre, called the pascal (Pa). The standard atmosphere is defined as exactly 101 325 Pa, and air at sea level is always close to it, so 1.0 × 10⁵ Pa is the working value; it is the one printed in the exam's table of information. That is a large number. On one face of a sheet of A4 paper, 0.210 m by 0.297 m, the air pushes with a force of

F = PA = (1.01 × 105 Pa)(0.210 m × 0.297 m) = (1.01 × 105 Pa)(0.0624 m2) = 6.3 × 103 N

which is the weight of about 640 kg. The paper does not fly off the desk, and is not pinned to it, because the air pushes on its other face with the same force. Pressure acts on both sides of anything immersed in a gas and in every direction at once. That is the first hint that pressure is not something that exists only at a wall.

One atom, one wall: the momentum bookkeeping

Here is the calculation that turns collisions into pressure, worked with real numbers first and symbols second.

Givens. A single helium atom of mass m = 6.65 × 10^-27 kg moves at v_x = 1000 m/s along the x-axis inside a cubical box of side L = 0.100 m. Its collisions with the walls are elastic.

The set-up in words. Look down on the box so that it appears as a square. The atom travels straight from the left wall to the right wall, bounces, travels back, bounces off the left wall, and returns. Watch only the right wall and ask what average force the atom exerts on it.

Principle. The impulse-momentum theorem: the average force on the atom multiplied by a time interval equals the atom's change in momentum during that interval. By Newton's third law the atom pushes on the wall with a force of equal size in the opposite direction.

1. Momentum change per collision. Before the bounce the atom's momentum is +mv_x; after it, -mv_x. The change has magnitude 2mvx = 2(6.65 × 10-27 kg)(1000 m/s) = 1.33 × 10-23 kg·m/s. The factor of 2 matters: an atom that bounces back pushes twice as hard as one that would merely stop.
2. Time between hits on the same wall. A round trip covers 2L, so Δt = 2L/vx = 2(0.100 m)/(1000 m/s) = 2.00 × 10-4 s, or 5000 hits per second.
3. Average force on the wall. F = 2mvx/Δt = (1.33 × 10-23 kg·m/s)/(2.00 × 10-4 s) = 6.65 × 10-20 N. In symbols, F = 2mvx/(2L/vx) = mvx2/L.

Sanity check. The units are kg·m/s divided by s, which is kg·m/s², a newton. The size is absurdly small, as it should be for one atom. A litre of air at room conditions holds about 2.5 × 10²² molecules, and a factor of that size turns 10^-20 N into a force you could feel.

The variant that flips the result. Now let the right wall move. Make it a piston pushed inward at 2.0 m/s. In the piston's own frame the atom approaches at 1002 m/s and, the collision being elastic, leaves at 1002 m/s; back in the frame of the room it leaves at 1004 m/s. Every atom that strikes the advancing piston comes away faster than it arrived, so a gas warms as it is compressed: the push on the piston does work on the gas. Draw the piston outward at 2.0 m/s instead and each atom leaves at 996 m/s, so a gas cools as it expands against a retreating piston. The same bookkeeping that gave the pressure predicts heating on compression and cooling on expansion, which is one reason the barrel of a bicycle pump grows warm as you use it. Lesson 4 turns this into the first law of thermodynamics.

Key idea: the force a gas exerts on a wall is the momentum delivered per collision times the number of collisions per second. Both factors grow with speed, so the force grows with the square of the speed.

From one atom to a whole gas

A real gas has N atoms moving in every direction at many speeds. For any one atom, the speed and its components are related by Pythagoras in three dimensions: v2 = vx2 + vy2 + vz2. If the directions are random, no axis is special, so on average each component carries a third of the total: (vx2)avg = (1/3)(v2)avg. Adding the forces of all N atoms on the right wall gives

F = Nm(v2)avg/(3L)

and dividing by the wall's area L², with L³ = V the volume of the box,

P = Nm(v2)avg/(3V), so PV = (1/3)Nm(v2)avg = (2/3)NKavg

where Kavg = (1/2)m(v2)avg is the average translational kinetic energy of one atom. A cube made the algebra easy, but the result holds for a container of any shape, because in equilibrium the pressure is the same at every point of the gas. Notice also what the derivation assumed, because these assumptions define an ideal gas: the atoms are so small that their own volume is negligible, their velocities are random, their collisions are elastic, and they exert no forces on one another except during collisions. Air at room temperature and pressure satisfies them well; a gas about to condense into a liquid does not.

The result explains the opening claim. Double the speed of every atom and each collision delivers twice the momentum, while collisions happen twice as often. The pressure goes up by a factor of four.

Temperature is the average kinetic energy

Measurements on real gases at low density, the subject of the next lesson, show that PV = NkBT, where T is the absolute temperature in kelvin and k_B is the Boltzmann constant. Since the 2019 revision of the SI, k_B has a fixed, exact value: 1.380 649 × 10^-23 J/K, which the exam's table rounds to 1.38 × 10^-23 J/K. Setting the kinetic result equal to the measured one, (2/3)NKavg = NkBT, and the number of atoms cancels:

Kavg = (3/2)kBT = (1/2)mvrms2

This is the microscopic meaning of temperature. The absolute temperature of a gas is proportional to the average kinetic energy of its atoms, and the proportionality holds only on a scale that starts at zero energy: T(K) = T(°C) + 273.15. The square root of the average squared speed is called the rms speed, short for root-mean-square:

vrms = √(3kBT/m)

Givens. Air at 20.0 °C, so T = 293 K. A nitrogen molecule (N₂) has m = 4.65 × 10^-26 kg; a helium atom has m = 6.65 × 10^-27 kg.

The set-up in words. Two flasks sit side by side on a bench, one of nitrogen and one of helium, both at room temperature. Find the average kinetic energy and the rms speed in each.

Principle. K_avg = (3/2)k_BT depends only on temperature; the rms speed then depends on the mass.

1. Kavg = (3/2)(1.38 × 10-23 J/K)(293 K) = 6.07 × 10-21 J, the same in both flasks.
2. Nitrogen: vrms = √[3(1.38 × 10-23 J/K)(293 K)/(4.65 × 10-26 kg)] = √(2.61 × 105 m2/s2) = 511 m/s.
3. Helium: vrms = √[3(1.38 × 10-23 J/K)(293 K)/(6.65 × 10-27 kg)] = √(1.82 × 106 m2/s2) = 1350 m/s.

Sanity check. A joule per kilogram is a square metre per square second, so the square root is a speed. The helium atom has a seventh of the nitrogen molecule's mass and the same kinetic energy, so it should be faster by √7 ≈ 2.65, and 511 m/s × 2.65 = 1350 m/s.

The variant that flips the result. Ask which flask has the greater pressure if both hold the same number of particles in the same volume. The helium atoms are 2.65 times faster, which tempts you to say helium. But pressure depends on m(v²)_avg, which is twice K_avg and therefore the same for both: the nitrogen molecules are slower but heavier, and each delivers a bigger momentum change per hit. The pressures are equal. For 2.50 × 10²² helium atoms in 1.00 L at 20 °C, P = Nm(v2)avg/(3V) = (2.50 × 1022)(6.65 × 10-27 kg)(1350 m/s)2/[3(1.00 × 10-3 m3)] = 1.01 × 105 Pa, one atmosphere, and the same number of nitrogen molecules gives the same answer.

Key idea: at a given temperature, every kind of gas particle has the same average kinetic energy. Lighter particles move faster, and the pressure does not care which kind you have, only how many there are per unit volume and how hot they are.

Not every molecule moves at 511 m/s

The rms speed is a kind of average, and an average hides a spread. Collisions constantly exchange energy, so at any moment some molecules are nearly at rest and a few are moving at several times the rms speed. The distribution of speeds was worked out by James Clerk Maxwell and Ludwig Boltzmann and is called the Maxwell-Boltzmann distribution. Picture its graph: speed on the horizontal axis, and on the vertical axis the fraction of molecules per unit speed interval. The curve starts at zero at zero speed, rises to a peak at the most probable speed, and falls away more slowly than it rose, leaving a long tail toward high speeds. The area under the curve represents all of the molecules, so it cannot change while the number of molecules stays fixed.

The exam does not ask for the formula of this curve, only for how its features respond to temperature, so the numbers below were computed from it for nitrogen.

Read the table across a row and down a column. In every row the most probable speed is below the rms speed, because the long tail pulls the rms value up. Going down, raising the temperature from 293 K to 600 K moves the peak to a higher speed by a factor of about 1.43, the square root of the temperature ratio. Because the area is fixed, a curve that spreads to the right must also become lower, so the peak flattens and the whole distribution broadens. The tail responds most dramatically of all: the fraction of molecules faster than 1000 m/s grows fourteen-fold. Any process that needs an unusually fast molecule, such as a water molecule escaping from the surface of a puddle, is far more sensitive to temperature than the average speed suggests.

Mass matters in the other direction. Helium at 293 K has a most probable speed of 1104 m/s, and 65 percent of its atoms are faster than 1000 m/s. At the same temperature, a lighter gas has a broader, flatter distribution that reaches much higher speeds.

Key idea: on a Maxwell-Boltzmann graph, a higher temperature or a lighter molecule moves the peak to the right and makes the curve lower and wider, with the same area underneath.

Pressure lives inside the gas, not only at the walls

The derivation used walls, but nothing in it needs them. Draw an imaginary square in the middle of a container. Molecules cross it in both directions all the time, carrying momentum with them, and a small pressure sensor placed there would be struck on its faces exactly as a wall is. Neglecting the weight of the gas itself, the sensor reads the same pressure anywhere in the container. Pressure is a property of the gas at every point.

That is why your body is not crushed by an atmosphere. Roughly 10⁵ Pa presses on every square metre of your skin, and the fluids and gases inside you press back with very nearly the same pressure. You notice the atmosphere only when the two sides disagree, as when your ears pop in a climbing aircraft: the cabin pressure falls, the air trapped behind each eardrum is briefly at a higher pressure than the air outside, and the eardrum bulges until a tube from the back of the throat lets the trapped air escape and the pressures equalise.

Common misconceptions

• "A hotter gas contains more molecules." Heating a sealed container leaves the number of molecules unchanged. The pressure rises because the same molecules move faster, so each impact is harder and impacts are more frequent.
• "At one temperature, every molecule has the same speed." The speeds are spread out in a Maxwell-Boltzmann distribution. At 20 °C about 41 percent of nitrogen molecules are faster than 500 m/s and a few are faster than 1000 m/s.
• "Heavier molecules carry more kinetic energy at the same temperature." The average kinetic energy is (3/2)k_BT whatever the molecule. Heavier molecules simply move more slowly.
• "Temperature is the total energy of the gas." It measures the average energy per molecule. A bathtub of water at 40 °C holds far more thermal energy than a cup of tea at 90 °C, though the tea is hotter.
• "Pressure exists only where gas touches a wall." Pressure exists throughout the gas, as the imaginary-square argument shows.
• "Going from 10 °C to 20 °C doubles the molecular energy." Kinetic energy is proportional to the absolute temperature, which rises from 283 K to 293 K, an increase of 3.5 percent. The next lesson is built around this trap.

Recap

A helium atom moving at 1000 m/s in a 0.100 m box reverses a momentum of 6.65 × 10^-24 kg·m/s five thousand times a second and pushes the wall with an average force of mv_x²/L, 6.65 × 10^-20 N. Summed over N atoms moving at random, that gives P = F/A = Nm(v²)_avg/(3V), so PV = (2/3)NK_avg, and a piston moving into the gas makes atoms rebound faster and warms it, while a retreating one cools it. Matching this to the measured law PV = Nk_BT shows that K_avg = (3/2)k_BT: 6.07 × 10^-21 J at 20 °C for any gas, which means 511 m/s for nitrogen and 1350 m/s for helium, and equal pressures for equal numbers of either. The speeds follow a Maxwell-Boltzmann distribution whose peak moves right and whose curve flattens and broadens as the temperature rises, with the fast tail growing fastest of all, and the pressure the molecules produce exists everywhere in the gas, not only at its walls.

Sources

1. OpenStax. (2022). 13.4 Kinetic theory: Atomic and molecular explanation of pressure and temperature. In College Physics 2e. openstax.org
2. OpenStax. (2022). 13.3 The ideal gas law. In College Physics 2e. openstax.org
3. Nave, R. (n.d.). Kinetic temperature, thermal energy; and Development of the Maxwell distribution. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
4. National Institute of Standards and Technology. (2024). CODATA 2022 values: Boltzmann constant (exact) and standard atmosphere (exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
5. National Institute of Standards and Technology. (n.d.). SI redefinition (in effect from 20 May 2019) and Kelvin (the Boltzmann constant fixed at exactly 1.380 649 × 10^-23 J/K). nist.gov; nist.gov (kelvin)
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 9.1, Kinetic theory of temperature and pressure. apcentral.collegeboard.org

A prediction that is wrong by a factor of fifteen

Suppose a student seals a glass jar of air at 20 °C and 101 kPa, then stands it in a water bath at 40 °C. Asked for the new pressure, she writes: "The temperature has doubled, so the pressure doubles, to 202 kPa." A pressure sensor wired through the lid reads 108 kPa. She predicted a rise of 101 kPa and the real rise was 7 kPa, so her answer is wrong by a factor of about fifteen, and yet each step looked sensible. The relationship she used, pressure proportional to temperature, is a genuine law. The mistake is in which temperature she fed into it.

This lesson finds that broken step, then a second trap that catches people who have fixed the first. Then it puts the repaired law to work: predicting, graphing, and finally measuring absolute zero from a graph that never reaches it.

Where the reasoning breaks: a zero that is not a zero

The previous lesson showed that the pressure of a gas comes from molecular impacts and that the average kinetic energy of the molecules is proportional to temperature, K_avg = (3/2)k_BT. That proportionality holds only on a scale whose zero is the temperature at which the molecular kinetic energy would vanish. The Celsius zero is not that point. It is the freezing point of water, chosen in the eighteenth century because it was easy to reproduce, and a gas at 0 °C is still full of molecules moving at hundreds of metres per second.

A quick absurdity test exposes the error. If pressure were proportional to Celsius temperature, a jar at 0 °C would exert no pressure at all, and a jar cooled to -10 °C would exert a negative pressure, pulling its walls inward. Neither happens. Converted to kelvin, 20 °C is 293 K and 40 °C is 313 K, and the ratio is 313/293 = 1.068. The molecules have 6.8 percent more kinetic energy, not twice as much, and the pressure rises 6.8 percent.

Remember: any equation in which temperature appears as a ratio or a product, such as PV = nRT, needs the absolute temperature in kelvin. A temperature difference is the same size in kelvin and in degrees Celsius, so only a change in temperature can be read straight off a Celsius thermometer.

The ideal gas law, in two forms

The full law relates all four quantities that describe a sample of gas:

PV = nRT = NkBT

P is the absolute pressure in pascals, V the volume in cubic metres and T the absolute temperature in kelvin. The two forms count the gas differently. N is the number of molecules and k_B the Boltzmann constant from Lesson 1. The amount n is measured in moles, where one mole contains the Avogadro number of particles, N_A = 6.022 140 76 × 10²³ per mole exactly, so n = N/N_A. The molar gas constant is R = N_Ak_B = 8.314 J/(mol·K), which the exam's table gives as 8.31 J/(mol·K). The units check: a pascal times a cubic metre is a newton per square metre times a cubic metre, which is a newton-metre, a joule, and R times kelvin times moles is also joules.

A quick calibration that is worth remembering: a 1.00 L bottle of air at 20 °C and 1.01 × 10⁵ Pa holds

n = PV/RT = (1.01 × 105 Pa)(1.00 × 10-3 m3)/[(8.31 J/(mol·K))(293 K)] = 0.0415 mol

which is 0.0415 × 6.02 × 10²³ = 2.50 × 10²² molecules, the figure used in Lesson 1. Turned around, one mole of any ideal gas at 0 °C and one atmosphere occupies 22.4 L. That volume is the same for hydrogen and for carbon dioxide: the law has no term for the kind of gas.

Worked problem: will the lid pop?

Givens. A rigid jar is sealed at 20 °C with air at 101 kPa inside and outside. Its lid is held on by a clip that releases if the inside pressure exceeds the outside pressure by more than 10 kPa. The jar is heated to 40 °C. The outside air stays at 101 kPa.

The set-up in words. A glass jar sits in a water bath. The amount of air inside is fixed because the jar is sealed, and its volume is fixed because glass barely expands. Only temperature and pressure can change.

Principle. With n and V constant, PV = nRT gives P/T = nR/V, a constant. So P₁/T₁ = P₂/T₂, with absolute pressure and absolute temperature.

Algebra. P2 = P1(T2/T1) = (101 kPa)(313 K/293 K) = 108 kPa. The excess over the outside air is 108 kPa - 101 kPa = 7 kPa, less than 10 kPa, so the lid holds.

Sanity check. The pressure rose 6.8 percent while the absolute temperature rose 6.8 percent, as proportionality requires. The size is reasonable: warming a jar of jam does not usually blow its lid off.

The variant that flips the result. Heat the jar to 60 °C instead. Now P2 = (101 kPa)(333 K/293 K) = 115 kPa, an excess of 14 kPa, and the lid releases. Notice also what the Celsius reasoning would have predicted: a pressure of 202 kPa at 40 °C, an excess of 101 kPa, and a lid that pops at a temperature where in fact it holds.

The second trap: what a tyre gauge actually reads

A tyre gauge reads gauge pressure, the amount by which the pressure inside exceeds the pressure of the atmosphere outside. A flat tyre reads zero on the gauge and yet is full of air at 101 kPa. The ideal gas law describes the molecules inside, all of them, so it needs the absolute pressure:

Pabsolute = Pgauge + Patmosphere

Suppose a tyre reads 220 kPa on the gauge on a 7 °C morning, and after a long drive the air inside is at 37 °C, with the volume essentially unchanged. A student who has learned the first lesson converts to kelvin but scales the gauge reading: 220 kPa × 310 K/280 K = 244 kPa. That is still wrong. The absolute pressure was 220 + 101 = 321 kPa, which scales to (321 kPa)(310 K/280 K) = 355 kPa, so the gauge now reads 355 - 101 = 254 kPa. The scaled-gauge answer is 10 kPa too low, because the atmosphere's share of the absolute pressure also rises with temperature. The absurdity test works here too: if the law used gauge pressure, a flat tyre at zero gauge would contain no air and could never be warmed into reading anything at all.

Four laws inside one

Hold two of the four quantities fixed and the ideal gas law reduces to a simpler law with its own name and its own graph. The graphs are described in words; sketching each one from its description is good practice for the exam.

A Boyle's law example with a sealed syringe: 20.0 mL of air at 101 kPa, compressed slowly so that it stays at room temperature, to 10.0 mL. Then P2 = P1V1/V2 = (101 kPa)(20.0 mL)/(10.0 mL) = 202 kPa. The litres cancel, which is one of the rare places where volume need not be converted to cubic metres. Push the plunger in quickly instead and the answer changes: the advancing piston speeds up the molecules, exactly as in Lesson 1, so the gas warms and the pressure briefly exceeds 202 kPa, settling back as the syringe cools to room temperature. Boyle's law needs the word "slowly".

Reading a graph to a temperature you can never reach

The pressure-temperature line gives a way to locate absolute zero without going there. A school apparatus for this is a sealed metal bulb of fixed volume connected to a pressure sensor. The bulb is placed in a sequence of water baths, each measured with a thermometer, and left until the reading stops changing. Here is an illustrative set of readings of the kind such an apparatus gives:

Plot pressure on the vertical axis against temperature in °C on the horizontal axis, choosing a horizontal scale that runs from about -300 °C to 100 °C so that there is room to extend the line. The six points lie close to a straight line. The best-fit line has a slope of 0.344 kPa per degree and meets the pressure axis at 94.5 kPa when the temperature is 0 °C. Extend it leftward to zero pressure:

0 = 94.5 kPa + (0.344 kPa/°C)T, so T = -94.5/0.344 = -275 °C

The accepted value of absolute zero is -273.15 °C, so this run is within 1 percent of the 368-degree span. Two features of the method matter for the exam. First, nobody follows the line all the way down: every real gas condenses into a liquid long before it gets there, so the graph is an extrapolation of ideal behaviour. Second, the extrapolation does not depend on the gas. Lines for different gases, or for the same gas at different starting pressures, have different slopes but meet the temperature axis at the same point, which is why that point deserves to be the zero of a temperature scale.

A design for this measurement should say four things:

• what is varied: the bath temperature, set with water baths and read on a thermometer;
• what is measured: the pressure, read when it is steady;
• what is controlled: the amount of gas and the volume of the bulb;
• how the data are analysed: a straight-line fit of P against T, extrapolated to P = 0.

Its weaknesses are worth naming too. The tube to the sensor sits at room temperature, so not all of the gas is at the bath temperature, and a scatter of only 0.3 kPa in the readings moves the intercept by several degrees.

Remember: a straight-line graph is the way to test a proportionality and to extract a constant from it. Here the slope tells you about the amount of gas and the intercept tells you where absolute zero is.

What makes a gas ideal, and when a real one is not

The ideal gas law is exact only for the model gas of Lesson 1: molecules with random velocities, whose own volume is negligible compared with the container, which collide elastically and exert no forces on each other except during collisions. Real gases at ordinary densities come close, because the molecules are far apart for most of the time. The model fails in two directions. At high pressure the molecules are packed closely enough that their own volume is no longer negligible. At low temperature, near the point where the gas condenses, the molecules move slowly enough that their mutual attractions matter, and the pressure falls below the ideal prediction. Steam just above its boiling point, or carbon dioxide compressed in a fire extinguisher, is far less ideal than the air in a classroom.

Common misconceptions

• "Pressure is proportional to temperature, so 40 °C gives twice the pressure of 20 °C." Only with absolute temperature: 313 K/293 K is a 6.8 percent rise. Celsius temperatures cannot be multiplied or divided.
• "The ideal gas law uses the reading on the gauge." It uses absolute pressure, gauge plus atmospheric. A flat tyre at zero gauge still holds air at 101 kPa.
• "Heavier gases exert more pressure." At the same n, V and T, every ideal gas has the same pressure. A litre of carbon dioxide and a litre of helium at the same temperature and pressure hold the same number of molecules.
• "Absolute zero was found by cooling a gas until its pressure disappeared." Every real gas liquefies first. Absolute zero is found by extrapolating the straight line, and the lines for all gases meet at the same temperature.
• "Boyle's law describes any compression." It describes a compression slow enough for the temperature to stay constant. A fast compression warms the gas and raises the pressure further.

What to carry forward

The student who predicted 202 kPa for a jar warmed from 20 °C to 40 °C fed Celsius temperatures into a proportionality that needs kelvin. The real pressure, 108 kPa, rises by the same 6.8 percent as the absolute temperature. The ideal gas law, PV = nRT = Nk_BT, needs absolute pressure as well as absolute temperature, which is why a tyre at 220 kPa gauge that warms from 7 °C to 37 °C ends at 254 kPa gauge and not 244. A litre of air at room conditions is 0.0415 mol, or 2.50 × 10²² molecules, and a mole of any ideal gas at 0 °C and one atmosphere fills 22.4 L. Holding two quantities fixed gives Boyle's, Charles's and Gay-Lussac's laws and Avogadro's principle, each with a graph you should be able to sketch.

The pressure-temperature line, extended to zero pressure, puts absolute zero near -273 °C for every gas, although no real gas survives that far as a gas. The model behind all of it assumes small, randomly moving, non-interacting molecules, and it fails at high pressures and near condensation.

Sources

1. OpenStax. (2022). 13.3 The ideal gas law. In College Physics 2e. openstax.org
2. OpenStax. (2022). 13.1 Temperature. In College Physics 2e. openstax.org
3. OpenStax. (2022). 11.6 Gauge pressure, absolute pressure, and pressure measurement. In College Physics 2e. openstax.org
4. National Institute of Standards and Technology. (2024). CODATA 2022 values: molar gas constant and Avogadro constant (both exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
5. National Institute of Standards and Technology. (n.d.). SI redefinition (the kilogram, kelvin, ampere and mole redefined in terms of constants of nature, in effect from 20 May 2019). nist.gov
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 9.2, The ideal gas law. apcentral.collegeboard.org

Two floors at 20 °C, and one bare foot

Step out of a shower onto a ceramic tile floor and it feels cold. Step onto the cotton bath mat beside it and the mat feels warm. Both have been in the same bathroom all night. Put a thermometer on each and both read 20 °C. The College Board lists this as one of the questions Unit 9 should be able to answer: why does the tile feel so much colder than the mat?

Two easy answers are wrong. The tile is not at a lower temperature, as the thermometer shows. And the tile does not "contain cold" that flows into your foot, because there is no such substance. Answering properly needs three ideas: what heating is, why energy moves from hot to cold, and what sets how fast it moves. The answer also predicts something you may not have noticed, which is that on a hot day the same tile feels hotter than the mat.

Heating is a transfer, not a substance

Two systems are in thermal contact when energy can pass between them by thermal processes, such as your foot pressed on a floor. Heating is the transfer of energy into a system by thermal processes, and cooling is the transfer of energy out of it. The energy transferred by heating is written Q, measured in joules, and it is positive when it enters the system. Physicists avoid saying an object "contains heat". An object contains internal energy, the kinetic and potential energy of its atoms; heat is the name for energy on its way from one object to another because their temperatures differ.

Temperature is something else again. It measures the average kinetic energy of the particles, as Lesson 1 showed, and it says which way energy will flow. Energy is transferred by thermal processes spontaneously from a higher-temperature system to a lower-temperature one, never the other way on its own. When no net energy flows between two systems in thermal contact, they are in thermal equilibrium, and that happens exactly when their temperatures are equal. This is why a thermometer works: it trades energy with whatever it touches until the two are in equilibrium, and then it reads its own temperature, which is now the same.

Why energy flows from hot to cold, one collision at a time

Picture the boundary between your foot and the floor at the scale of atoms. The atoms in your skin, at about 33 °C, jostle with a slightly higher average kinetic energy than the atoms of the tile at 20 °C. When a skin atom and a tile atom collide, energy can go either way in any single collision, but on average it goes from the more energetic atom to the less energetic one, just as a fast billiard ball that strikes a slow one usually slows down while the slow one speeds up. Over billions of collisions per second, the tile gains energy and your skin loses it.

Nothing forbids an individual collision from carrying energy the other way. The flow from hot to cold is a matter of overwhelming probability: after many collisions, the most probable state of the two systems by far is the one in which they share the same temperature. Lesson 5 builds the second law of thermodynamics on exactly this idea.

Three routes: conduction, convection and radiation

• Conduction is transfer through a material by collisions between neighbouring particles, with no bulk movement of the material. Your foot on the tile, a spoon warming from the soup end to the handle, and energy leaking through a window pane are conduction.
• Convection is transfer by the bulk motion of a fluid. Air warmed by a radiator expands, becomes less dense and rises, carrying its energy across the room while cooler air sinks to replace it.
• Radiation is transfer by electromagnetic waves, which need no material at all. Sunlight warms your face after crossing 150 million kilometres of nearly empty space. Every object emits radiation, more strongly the hotter it is, and Lesson 26 puts a number on that.

A cup of tea left on a desk loses energy by all three at once: conduction into the desk, convection as warm air and vapour rise from its surface, and radiation from its sides. For the bare foot on the floor, conduction dominates, so that is the route to quantify.

Conduction, quantified

Consider a slab of material of thickness L and face area A, with one face held at a higher temperature than the other by ΔT. The rate at which energy is conducted through it is

Q/Δt = kAΔT/L

The rate is in watts, joules per second. Each factor makes physical sense. A larger area offers more parallel paths. A larger temperature difference drives a larger flow. A thicker slab means the energy has to be passed along a longer chain of collisions. The constant k, the thermal conductivity, is a property of the material itself and depends on how its atoms are arranged and how they interact; it has units of W/(m·°C), the same as W/(m·K). Metals conduct well because, besides the vibrating atoms, they have free electrons that carry energy quickly. Here are values from OpenStax's table, measured near 0 °C:

The right-hand column is the story of insulation. Wool and down conduct almost as poorly as air, because they are mostly air, held still in small pockets so that it cannot form convection currents.

Givens. A foot with skin at 33 °C touches a floor at 20 °C over a contact area of A = 0.015 m². Model the tile as a 1.0 cm slab of fired clay with k = 0.84 W/(m·°C), the value listed for concrete brick, and the mat as a 1.0 cm layer with the conductivity of wool, 0.04 W/(m·°C). Treat the far side of each layer as staying at 20 °C.

The set-up in words. The sole of the foot sits on top of a thin layer, with energy flowing downward from the warm skin, through the layer, into the cooler floor beneath.

Principle. The conduction rate Q/Δt = kAΔT/L, with ΔT = 33 °C - 20 °C = 13 °C.

Algebra. Tile: Q/Δt = (0.84 W/(m·°C))(0.015 m2)(13 °C)/(0.010 m) = 16 W. Mat: Q/Δt = (0.04 W/(m·°C))(0.015 m2)(13 °C)/(0.010 m) = 0.78 W.

Sanity check. Watts come out, because W/(m·°C) times m² times °C divided by m is W. The ratio of the two rates is 0.84/0.04 = 21, simply the ratio of the conductivities, because everything else is the same. The model exaggerates both rates, since the skin touching a tile cools within moments and the temperature difference shrinks, but it gets the comparison right: the tile pulls energy out of your foot about twenty times faster than the mat. Your nerves sense the falling temperature of your own skin, not the temperature of the floor, and that is the coldness you feel.

The variant that flips the result. Now take the same tile and the same mat outdoors on a summer afternoon, both at 45 °C in the sun. The temperature difference is 45 °C - 33 °C = 12 °C, but now the floor is the hotter side, so energy flows into the foot: about 15 W through the tile and 0.72 W through the mat. The tile now feels hotter. Nothing about either material has changed. The same high conductivity that made the tile feel colder indoors makes it feel hotter outdoors, because the direction of the flow has reversed.

Worth holding on to: touch measures how fast energy enters or leaves your skin, not the temperature of what you touch. High-conductivity materials feel extreme in both directions.

Specific heat: how much energy a degree costs

Conduction tells you how fast energy moves. How much a given amount of energy changes an object's temperature depends on its mass and on what it is made of:

Q = mcΔT

Here m is the mass and c the specific heat, the energy needed to raise 1 kg of the material by 1 °C, in J/(kg·°C). Like thermal conductivity, specific heat is an intrinsic property that depends on the arrangement and interactions of the atoms. The exam treats it as independent of temperature, which is a good approximation for solids and liquids over everyday ranges. Water's value, 4186 J/(kg·°C), is one of the largest of any common substance, which is why a lake changes temperature far less over a day than the air above it does.

Warming 0.250 kg of water for a cup of tea from 20.0 °C to 100.0 °C takes Q = (0.250 kg)(4186 J/(kg·°C))(80.0 °C) = 8.37 × 104 J, which a 2.0 kW kettle supplies in about 42 s if none is wasted. The same 8.37 × 10⁴ J delivered to 0.250 kg of aluminium, with c = 900 J/(kg·°C), would raise it by 372 °C. Water is hard to warm and slow to cool.

Why should materials differ so much? Multiply each specific heat by the mass of one mole of the material, and a pattern appears for the metals:

Per kilogram, aluminium needs seven times the energy lead does. Per mole, which means per fixed number of atoms, they nearly agree, at about 25 J/(mol·K). A kilogram of aluminium simply contains far more atoms than a kilogram of lead, and at room temperature each atom of a metal stores about the same energy per degree, an observation known as the law of Dulong and Petit. Water, at 75 J/(mol·K), stores about three times as much per molecule as a metal stores per atom, because its molecules interact strongly with their neighbours and have more ways to hold energy. The specific heat is a window onto the atoms.

Worked problem: a hot block in cold water

Givens. A 0.200 kg aluminium block at 90.0 °C is dropped into 0.500 kg of water at 20.0 °C in an insulated cup. Specific heats: aluminium 900 J/(kg·°C), water 4186 J/(kg·°C). Neglect the cup.

The set-up in words. The block sinks to the bottom of the cup. Energy flows from the block into the water by conduction until both reach one final temperature T. The insulation means no energy enters or leaves the block-and-water system.

Principle. Conservation of energy for an isolated system: the energy lost by the block equals the energy gained by the water, each given by Q = mcΔT.

Algebra. (0.200 kg)(900 J/(kg·°C))(90.0 °C - T) = (0.500 kg)(4186 J/(kg·°C))(T - 20.0 °C). The products m c are the heat capacities, 180 J/°C for the block and 2093 J/°C for the water, so 180(90.0 - T) = 2093(T - 20.0), giving 16 200 - 180T = 2093T - 41 860, so 2273T = 58 060 and T = 25.5 °C.

Sanity check. The final temperature lies between the two starting temperatures, as it must. It sits much closer to the water's, because the water's heat capacity is more than eleven times the block's. Check the energy balance: the block gives up 180 × 64.5 = 11 600 J and the water gains 2093 × 5.5 = 11 500 J, equal within rounding.

The variant that flips the result. Reverse the roles: pour 0.200 kg of water at 90.0 °C into a 0.500 kg aluminium pan at 20.0 °C. Now the heat capacities are 837 J/°C for the hot water and 450 J/°C for the cold pan, and T = (837 × 90.0 + 450 × 20.0)/(837 + 450) = 65.5 °C. The final temperature now lands near the hot starting temperature. Which end it lands near is decided by which object has the larger heat capacity, not by which one is hot.

Worth holding on to: in a mixing problem, write "energy lost by the hot object equals energy gained by the cold one", with each temperature change written as a positive number, and expect the final temperature to sit near the object with the larger mc.

Common misconceptions

• "Metal objects are colder than wooden ones in the same room." They are at the same temperature. Metal conducts energy out of your skin hundreds of times faster, so your skin cools faster.
• "Cold flows into your hand." There is no substance called cold. Energy flows out of your hand into the colder object.
• "Heat and temperature are the same thing." Temperature is a property of a system; heat is energy transferred because of a temperature difference. The same 8.37 × 10⁴ J warms tea water by 80 °C and the same mass of aluminium by 372 °C.
• "A blanket or a jacket produces warmth." It produces nothing. It traps still air, which slows the rate at which your body's own energy escapes.
• "When two objects touch, the final temperature is the average of the two." Only when their heat capacities are equal. The aluminium block in water ends at 25.5 °C, not 55 °C.

What to remember

The tile and the mat in the bathroom are both at 20 °C, and the tile feels colder because its thermal conductivity, about 0.84 W/(m·°C) against 0.04 for a woollen mat, lets it draw energy out of your skin about twenty times faster; on a 45 °C patio the same tile feels hotter, because the flow reverses. Heating and cooling are transfers of energy, by conduction, convection or radiation, and the transfer runs spontaneously from higher to lower temperature because collisions on average pass energy from faster atoms to slower ones until thermal equilibrium makes the temperatures equal. The rate of conduction is Q/Δt = kAΔT/L. The energy to change a temperature is Q = mcΔT, 8.37 × 10⁴ J to boil a cup of water from room temperature, and specific heat reflects the atoms themselves: per mole, metals from aluminium to lead need about the same 25 J/(mol·K). In a mixing problem energy lost equals energy gained, and the final temperature lands near whichever object has the larger heat capacity, 25.5 °C for a hot block in a cup of water and 65.5 °C for hot water in a cold pan.

Sources

1. OpenStax. (2022). 14.5 Conduction. In College Physics 2e (Table 14.3, thermal conductivities). openstax.org
2. OpenStax. (2022). 14.2 Temperature change and heat capacity. In College Physics 2e (Table 14.1, specific heats). openstax.org
3. OpenStax. (2022). 14.4 Heat transfer methods. In College Physics 2e. openstax.org
4. Nave, R. (n.d.). Law of Dulong and Petit. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 9.3 and 9.5, and the Unit 9 essential questions. apcentral.collegeboard.org

Same start, same finish, two different bills

Two identical cylinders each hold a monatomic gas at 100 kPa in a volume of 1.00 L. Both end at 200 kPa and 2.00 L. The first cylinder gets there by expanding at constant pressure and then being heated with its piston locked; the second has its piston locked first and expands afterwards. Measure the energy that entered each one by heating and you find 550 J for the first and 650 J for the second. Same starting state, same final state, and a difference of 100 J in what it cost.

The difference is not a measurement error. It is the most important fact about thermodynamic processes: heating and work depend on the route a gas takes, while its internal energy depends only on where it is. This lesson builds the tool that keeps the accounts straight, the first law of thermodynamics drawn on a pressure-volume diagram, and then lays four standard processes side by side so that you can see what each one does with the energy.

What a gas stores: internal energy

The internal energy U of a system is the total kinetic energy of the particles that make it up plus the potential energy of their configuration. In an ideal gas the atoms exert no forces on each other between collisions, so there is no potential energy to count, and for a monatomic gas, the kind the exam assumes unless told otherwise, the internal energy is just the total translational kinetic energy of the atoms. Lesson 1 gave the average per atom as (3/2)k_BT, so

U = (3/2)NkBT = (3/2)nRT = (3/2)PV

The last form uses PV = nRT and is handy on a PV diagram: the internal energy can be read straight off the coordinates. For one mole of helium at 300 K, U = (3/2)(8.31 J/(mol·K))(300 K) = 3740 J. A useful unit fact follows from the last form: a kilopascal times a litre is 10³ Pa times 10^-3 m³, exactly one joule, so PV in kPa·L is already in joules.

Internal energy belongs to the inside of a system, not to its motion as a whole. A sealed box of gas carried along on a moving train has a large kinetic energy as a box, but heating the gas raises its internal energy without changing the motion of the box's centre of mass at all, and setting the box moving does not warm the gas. Because U depends only on the state of the gas, on its temperature for a given amount of ideal gas, the change ΔU between two states is the same whatever route connects them. In the opening example, U went from (3/2)(100 kPa)(1.00 L) = 150 J to (3/2)(200 kPa)(2.00 L) = 600 J in both cylinders, a rise of 450 J each.

The first law, and the sign that trips everyone

Energy can enter a gas in two ways: by heating, when it is in thermal contact with something hotter, and by work, when a piston pushes on it. The first law of thermodynamics is conservation of energy written for exactly that situation:

ΔU = Q + W

Q is the energy transferred into the gas by heating, positive when energy enters. W is the work done on the gas, positive when the gas is compressed. For a constant or average external pressure acting through a volume change,

W = -PΔV

The minus sign does the bookkeeping: compression makes ΔV negative and W positive, adding energy to the gas, while expansion makes W negative, because the gas is doing work on its surroundings and paying for it out of its own energy. This is the convention printed in the exam's table of information, where W is defined as the work done on a system. Be warned that many textbooks, OpenStax College Physics among them, write the first law as ΔU = Q - W with W the work done by the system. The physics is identical and the sign of W is opposite, so always check which W a source means before copying its equations.

The point: ΔU depends only on the end states; Q and W separately depend on the path. The first law says that whatever route is taken, they must add up to the same ΔU.

Work is an area on the PV diagram

A PV diagram plots the state of a gas as a point, with volume on the horizontal axis and pressure on the vertical axis, and a process as a path from one point to another. At constant pressure the path is a horizontal line, and the work done on the gas is -PΔV, minus the area of the rectangle beneath the line. When the pressure varies, the rule generalises: the magnitude of the work done on the gas equals the area under the path, between it and the volume axis. The sign comes from the direction. A path running to the right, toward larger volume, is an expansion and the work done on the gas is negative; a path running to the left is a compression and the work done on the gas is positive.

The opening puzzle now reads off the diagram. Path 1 runs from (1.00 L, 100 kPa) right to (2.00 L, 100 kPa), then straight up to (2.00 L, 200 kPa). The area under it is the rectangle 100 kPa by 1.00 L, so W = -100 J, and Q = ΔU - W = 450 J - (-100 J) = 550 J. Path 2 runs straight up first to (1.00 L, 200 kPa), then right at 200 kPa. The area under it is twice as big, so W = -200 J and Q = 450 J + 200 J = 650 J. The second gas did 100 J more work on its surroundings, so it needed 100 J more energy from heating to end up in the same state.

Curves of constant temperature on a PV diagram are called isotherms. For an ideal gas each is a curve of PV = constant, falling toward both axes without reaching them, and isotherms farther from the origin are hotter. The opening gas ended at four times its starting value of PV, so its final state lies on the isotherm of four times the starting absolute temperature.

Four processes side by side

Four special processes appear again and again, each defined by one thing held fixed. Here they are with what the first law says about each, for a monatomic ideal gas.

Read down the ΔU column first, because it is where the physics of each row lives. Only the isothermal row has ΔU = 0, since only there is the temperature fixed. Then read across. In the isovolumetric row no work is possible, so heating goes entirely into internal energy. In the isothermal row the internal energy cannot change, so every joule of work done on the gas must leave again by cooling. In the adiabatic row heating is shut off, so work is the only way energy gets in or out, and the temperature has to move.

Reading the table, one row at a time

Givens. A cylinder of helium under a freely sliding piston stays at 1.00 × 10⁵ Pa while it is heated and its volume grows from 2.00 L to 3.00 L.

The set-up in words. A vertical cylinder with a weighted piston on top, sitting on a hot plate. The weight fixes the pressure; as the gas warms, the piston rises.

Principle. The first law, with W = -PΔV at constant pressure and U = (3/2)PV.

1. Work on the gas: W = -PΔV = -(1.00 × 105 Pa)(3.00 - 2.00) × 10-3 m3 = -100 J.
2. Change in internal energy: ΔU = (3/2)PΔV = (3/2)(100 kPa)(1.00 L) = 150 J.
3. Heating: Q = ΔU - W = 150 J - (-100 J) = 250 J.

Sanity check. The gas expanded, so it did work on the piston and W is negative. The temperature rose, since PV went up from 200 J to 300 J, so ΔU is positive. Of the 250 J delivered by the hot plate, 150 J stayed as faster atoms and 100 J lifted the piston, and the split of 3 to 2 is exactly the ratio of (3/2) to 1 in the table.

The variant that flips the result. Run the same gas backwards at the same pressure, from 3.00 L to 2.00 L. Now W = +100 J, because the piston does work on the gas, and ΔU = -150 J, because PV falls. The first law gives Q = -150 J - 100 J = -250 J. Work is being done on the gas and yet energy must be removed by cooling, 250 J of it. A constant-pressure compression is impossible unless the gas is chilled at the same time.

The other three rows, more briefly. Isovolumetric: a rigid 2.00 L tank heated from 100 kPa to 150 kPa has W = 0 and Q = ΔU = (3/2)VΔP = (3/2)(2.00 L)(50 kPa) = 150 J. Isothermal: compress 2.00 L of gas at 100 kPa slowly to 1.00 L in a water bath. The pressure climbs along the isotherm to 200 kPa, and the work done on the gas is the area under that curve. A straight line between the endpoints would enclose a trapezium of 150 J; the isotherm sags below that line, and the area under it works out at about 139 J. Since ΔU = 0, Q = -139 J: the bath carries away exactly the energy the piston puts in. Adiabatic: compress the same gas so fast that no energy can leave by heating. Then Q = 0 and ΔU = W, which is positive, so the gas warms; its pressure therefore climbs faster than along the isotherm, which is why an adiabat is steeper. Run it in reverse and an expanding gas with no heating supply cools, which is why gas released from a pressurised cylinder comes out noticeably cold.

A cycle, and the work hidden in the loop

An engine takes a gas around a closed loop on the PV diagram again and again. Because the gas returns to its starting state, ΔU = 0 for a full cycle, and the first law says the net heating must equal minus the net work done on the gas. The net work done by the gas is the area enclosed by the loop, positive when the loop runs clockwise. Here is a rectangular cycle through A (1.00 L, 300 kPa), B (3.00 L, 300 kPa), C (3.00 L, 100 kPa) and D (1.00 L, 100 kPa), worked leg by leg with U = (3/2)PV.

The totals check three ways. ΔU sums to zero, as it must for a return to the starting state. The net work done by the gas, 400 J, equals the enclosed area, 200 kPa by 2.00 L. And the heating column shows that the gas took in 1500 J + 300 J = 1800 J on the legs where it was heated and gave out 900 J + 500 J = 1400 J where it was cooled. Of every 1800 J supplied, 400 J came out as work, about 22 percent, and 1400 J had to be dumped. Why an engine cannot avoid that dumping is the subject of the next lesson. Notice too that A and C sit on the same isotherm, since PV is 300 J at both.

The point: on a PV diagram, area under a path is work, a clockwise loop is an engine, and the first law turns each leg's work and change in PV into the heating that leg needs.

Common misconceptions

• "Isothermal means no heat is transferred." That is adiabatic. Isothermal means the temperature, and so U, is constant, which for a compression requires energy to leave by cooling at exactly the rate work brings it in.
• "Adding energy by heating always raises the temperature." In an isothermal expansion, heating supplies energy that leaves at once as work, and the temperature does not change.
• "Compressing a gas always warms it." A slow compression in a water bath keeps it at the bath's temperature; a fast compression with no time for cooling warms it.
• "W in the first law is the work the gas does." In the exam's convention W is the work done on the gas, so ΔU = Q + W. With the other convention the equation becomes ΔU = Q - W. Mixing the two flips the sign of the answer.
• "A gas contains a certain amount of heat and a certain amount of work." It contains internal energy. Heat and work are transfers that happen during a process, and how much of each depends on the path.

The short version

Two cylinders taken from 100 kPa and 1.00 L to 200 kPa and 2.00 L by different routes both gain 450 J of internal energy, but one needs 550 J of heating and the other 650 J, because they do different amounts of work. Internal energy is a property of the state, U = (3/2)nRT = (3/2)PV for a monatomic ideal gas, while heating and work depend on the path. The first law in the exam's convention is ΔU = Q + W, with W = -PΔV the work done on the gas, positive for compression; OpenStax's ΔU = Q - W counts work the other way. On a PV diagram the work is minus the area under an expanding path. The four standard processes each fix one thing: isovolumetric (W = 0, Q = ΔU), isobaric (a 2.00 L to 3.00 L expansion at 100 kPa takes 250 J, 150 J to internal energy and 100 J to work), isothermal (ΔU = 0, Q = -W) and adiabatic (Q = 0, ΔU = W, so compression heats and expansion cools). Round a closed loop, ΔU = 0 and the enclosed area is the net work: the rectangular cycle turns 1800 J of heating into 400 J of work and rejects 1400 J.

Sources

1. OpenStax. (2022). 15.1 The first law of thermodynamics. In College Physics 2e (note: written as ΔU = Q - W with W done by the system). openstax.org
2. OpenStax. (2022). 15.2 The first law of thermodynamics and some simple processes. In College Physics 2e. openstax.org
3. OpenStax. (2022). 14.1 Heat (Joule's experiment and the mechanical equivalent of heat). In College Physics 2e. openstax.org
4. Nave, R. (n.d.). First law of thermodynamics. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 9.4, and Table of Information: Equations (W defined as work done on a system). apcentral.collegeboard.org

A bead 6.3 micrometres across

In July 2002, Genmiao Wang, Edith Sevick, Emil Mittag, Debra Searles and Denis Evans reported in Physical Review Letters that they had watched the second law of thermodynamics run backwards. They held latex beads 6.3 micrometres across, each in the focus of a laser beam that acts as a trap, in a cell of water, and moved the trap through the water. Dragging a bead through water normally turns work into thermal energy of the water, the way stirring tea warms it slightly. But over stretches of about a tenth of a second, some of the beads' trajectories did the opposite: the jostling water molecules pushed the bead along, and thermal energy was turned into work. The share of such trajectories fell as the stretches got longer, and over about two seconds none were seen. The title of the paper called them violations of the second law. Within days, Theo Nieuwenhuizen and Armen Allahverdyan posted a reply arguing that the result was not a violation but even a confirmation of that law.

Both sides were serious physicists looking at the same data, so the disagreement is worth understanding. It turns on what the second law actually claims. This lesson states the law and shows where its force comes from. It takes two challenges to it seriously, an everyday one and this modern one. On the way it explains why no engine can ever turn all of its fuel's energy into work.

What the second law actually claims

The first law says energy is conserved. It does not say which way things go: a cup of tea that spontaneously grew hotter by drawing energy from the cooler air around it would conserve energy perfectly. Yet tea always cools. The second law supplies the direction, and it comes in three equivalent forms.

• About heat flow: energy is transferred by heating spontaneously from a hotter body to a colder one, and never spontaneously in the reverse direction.
• About engines: no engine working in a cycle can take energy from a single reservoir by heating and convert all of it into work.
• About entropy: the total entropy of an isolated system never decreases. It stays constant only if every process in the system is a reversible process, one that could be run backwards leaving no trace anywhere, and no real process is.

The exam treats entropy qualitatively, and the College Board describes it in two ways. Entropy measures how spread out a system's energy is: concentrated energy, such as the energy of a hot coal in a cold room, tends to disperse until it is shared evenly. And entropy measures how much of a system's energy is unavailable to do work: once the coal's energy has spread through the room at one temperature, no engine can extract work from it. Entropy is also a state function, like internal energy: it depends only on the current state of a system, not on how the system got there. An isolated system drifts spontaneously toward thermodynamic equilibrium, and at equilibrium its entropy is as large as it can be.

Why spreading wins: counting the arrangements

Why should energy and matter spread rather than gather? Not because any law of motion pushes them outward. Collisions run equally well forwards and backwards. The answer is counting. Put four gas molecules in a box and ask how likely it is that, at some instant, all four are in the left half. Each is equally likely to be in either half, so the chance is (1/2)⁴ = 1/16. You would see it happen regularly. Now increase the number of molecules.

OpenStax makes the same point with 100 coins. There is exactly one way for all 100 to land heads, and about 1.0 × 10²⁹ ways to get 50 heads and 50 tails, out of 1.27 × 10³⁰ possible outcomes in all. Every individual arrangement is equally likely; there are simply vastly more spread-out arrangements than gathered ones. Energy behaves the same way. There are overwhelmingly more ways to share energy evenly among the atoms of a hot coal and a cold room than to keep it piled up in the coal, so collisions, shuffling at random, carry the system toward the even share and keep it there. That is what the drift to equilibrium means, and it is why the second law is, at bottom, a law of probability. For four molecules the probability is weak. For anything you can see, it is so strong that a decrease of entropy is never observed.

An optional number makes the heat-flow version concrete. For energy Q transferred by heating at a steady absolute temperature T, the entropy change is Q/T; the exam does not ask for this, but it shows why heat runs downhill. Move 1000 J from a body at 400 K to one at 300 K. The hot body loses 1000/400 = 2.50 J/K and the cold one gains 1000/300 = 3.33 J/K, so the total rises by 0.83 J/K. The reverse transfer would lower the total by 0.83 J/K, which an isolated pair of bodies never does.

The everyday objection: a freezer that makes order

Here is the first challenge, and it is a fair one. Put a tray of water in a freezer and a few hours later the molecules, which were tumbling freely, are locked into the ordered lattice of ice. Their entropy has gone down. Living things do the same all day long, building large, ordered molecules out of small, scattered ones. If entropy never decreases, how can any of this happen?

The answer is in two words of the law's statement: isolated system. The water in the freezer is not isolated. It is a closed system, which means no matter crosses its boundary but energy does, and energy is leaving it. The College Board's wording is exact: the entropy of an isolated system never decreases, but the entropy of a closed system can decrease, because energy can be transferred into or out of it. To test the second law you must draw the boundary around everything involved: the water, the freezer, its motor, and the kitchen it warms.

Do that with numbers. Suppose the freezer removes 1000 J from its contents at 255 K, using 500 J of electrical work to do it. The contents lose 1000/255 = 3.92 J/K of entropy. The freezer must deliver the 1000 J it removed plus the 500 J of work to the kitchen, 1500 J in all, through the coils at its back. The kitchen at 295 K gains 1500/295 = 5.08 J/K. The total rises by 1.16 J/K. The ice became more ordered and the universe as a whole did not. Living things are the same: their order is paid for by the much larger disorder they create in their surroundings, which is why they need food and why they give off heat.

In short: entropy can fall inside a boundary that energy is crossing. Draw the boundary wide enough that nothing crosses it, and the total never falls.

Heat engines, and the energy they must throw away

A heat engine takes energy Q_H by heating from a hot reservoir, converts part of it into work W, and rejects the rest, Q_C, by heating a cold reservoir, returning to its starting state each cycle. Since ΔU = 0 for a full cycle, energy conservation gives Q_H = W + Q_C. The engine's efficiency is the fraction of the input that comes out as work:

e = W/QH = 1 - QC/QH

The engine form of the second law says Q_C can never be zero, so e is always less than 1. The cycle analysed in Lesson 4 took in 1800 J, did 400 J of work and rejected 1400 J, for an efficiency of 22 percent.

Givens. A gas-fired power station takes energy from burning fuel at a rate of 2.40 × 10⁹ W and produces 1.00 × 10⁹ W of electrical power. Its steam reaches the turbines at 810 K, and its cooling water is at 300 K.

The set-up in words. Fuel burns under a boiler; steam spins a turbine connected to a generator; the spent steam is condensed by cooling water drawn from a river, which carries the rejected energy away.

Principle. Energy conservation for a cyclic engine, Q_H = W + Q_C, applied to rates, and the definition e = W/Q_H.

1. Efficiency: e = (1.00 × 109 W)/(2.40 × 109 W) = 0.417, about 42 percent.
2. Rejected power: QC/Δt = (2.40 - 1.00) × 109 W = 1.40 × 109 W, sent into the river and the air.

Sanity check. OpenStax quotes about 42 percent as typical for stations fired by coal, oil and gas, and about 35 percent for nuclear stations, so the figure is realistic. A deeper check comes from the best efficiency any engine could have between these two temperatures. The French engineer Sadi Carnot worked it out in 1824 for an ideal reversible engine: emax = 1 - TC/TH = 1 - (300 K)/(810 K) = 0.63. That formula is not on the exam's equation sheet, but it explains why power stations run their steam as hot as their materials allow. The station's 42 percent is well below the 63 percent ceiling, as every real engine is.

The variant that flips the result. Now consider the ship from the second law's engine form: a design that takes energy from the warm sea and turns it into work to drive the propellers, with no colder reservoir at all. Put T_C = T_H into the same ceiling and the maximum efficiency becomes 1 - 1 = 0. With one reservoir, the work is zero however clever the machinery. The first law would happily allow the ship; the second law sinks it, because energy spread evenly through the sea at one temperature is exactly the energy that is unavailable to do work.

The modern objection: violation or confirmation?

Back to the bead. Wang and colleagues saw real entropy-consuming stretches, and their evidence is clear. Over about a tenth of a second some trajectories turned thermal energy of the water into work. The fraction fell as the time window grew, and none appeared over windows of about two seconds. If the second law is a statement that every process in every system, however small and brief, produces entropy, then the bead broke it, and the paper's title is accurate.

Nieuwenhuizen and Allahverdyan read the law differently. On their reading the second law is a statement about the behaviour of systems taken as a whole: about averages, and about large systems over significant times. The bead's entropy production, averaged over many trajectories, was positive, exactly as the law requires, and the rate at which the exceptions disappeared with time is what the statistical theory behind the experiment, called the fluctuation theorem, predicts. On that reading the experiment confirms the law and measures how it emerges.

What would settle it is not another measurement, since nobody disputes the data, but agreement on the wording. The counting argument above tells you which wording physicists mostly use in practice. For four molecules the second law is a weak tendency; for the 2.5 × 10²² molecules in a bottle of air it is a certainty for every practical purpose. The bead was small enough, and the windows short enough, to see the edge where one becomes the other. For anything you can hold, from a cup of tea to a power station, entropy goes up.

Common misconceptions

• "Entropy can never decrease anywhere." It can decrease in a closed system that loses energy, such as water freezing in a freezer. Only the total for an isolated system cannot decrease.
• "A refrigerator breaks the second law by making heat flow from cold to hot." It does so only because work is supplied, and the room gains more entropy than the cold interior loses.
• "A perfect engine is just a matter of better engineering." The second law forbids any cyclic engine from converting all its heat input to work; even an ideal reversible engine between 810 K and 300 K tops out at 63 percent.
• "If energy is conserved, the process can happen." Energy conservation is necessary but not sufficient. Tea warming itself from cooler air would conserve energy and never happens.
• "Entropy depends on how a system got to its state." Entropy is a state function. A gas that expands freely into a vacuum and one that expands slowly against a piston, ending in the same state, have the same entropy change.

Where this leaves us

In 2002 a 6.3 micrometre bead in an optical trap turned thermal energy into work for stretches of about a tenth of a second, never over about two seconds. Two groups of physicists disagreed about whether that violated the second law or confirmed it. The disagreement is about whether the law describes every tiny trajectory or the overwhelming behaviour of systems large enough to matter.

The law says that energy flows by heating from hot to cold and that no cyclic engine turns all its heat into work. It also says that the entropy of an isolated system never decreases. Entropy means how spread out energy is, and how unavailable it is for work. Its force is counting: four molecules gather in half a box one time in sixteen, a litre of air essentially never. Entropy can fall inside a closed system such as a freezer. The freezer removes 1000 J from its contents and dumps 1500 J into the kitchen, so the total still rises by 1.16 J/K. Every heat engine obeys Q_H = W + Q_C with Q_C greater than zero. A gas-fired station turning 2.40 GW into 1.00 GW is 42 percent efficient and must reject 1.40 GW, below Carnot's ceiling of 63 percent. A ship trying to run on the warmth of the sea alone could do no work at all.

Sources

1. Wang, G. M., Sevick, E. M., Mittag, E., Searles, D. J., and Evans, D. J. (2002). Experimental demonstration of violations of the second law of thermodynamics for small systems and short time scales. Physical Review Letters, 89(5), 050601. https://doi.org/10.1103/PhysRevLett.89.050601. Record and abstract: digitalcollections.anu.edu.au
2. Nieuwenhuizen, T. M., and Allahverdyan, A. E. (2002). Comment on "Experimental violations of the second law of thermodynamics for small systems and short timescales". arXiv:cond-mat/0207587. arxiv.org
3. Physics World. (2002, July 16). Small systems defy second law (bead size and time scales of the 2002 experiment). physicsworld.com
4. OpenStax. (2022). 15.3 Introduction to the second law of thermodynamics: Heat engines and their efficiency; 15.4 Carnot's perfect heat engine. In College Physics 2e. openstax.org
5. OpenStax. (2022). 15.7 Statistical interpretation of entropy and the second law of thermodynamics. In College Physics 2e. openstax.org
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 9.6, Entropy and the second law of thermodynamics (qualitative treatment only). apcentral.collegeboard.org

Two strips of tape that will not touch

Press a 15 cm strip of clear sticky tape onto a tabletop, fold one end under to make a handle, and rip it off quickly. Do the same with a second strip, then hang both from the edge of a shelf a few centimetres apart. They swing away from each other. Bring your finger near either one and it swings toward your finger. Now make a different pair: press one strip on the table, press a second strip on top of the first, and peel the two off the table together. Pull them apart by their handles and hang them up. These two swing toward each other. Each also swings toward your finger, and toward a scrap of paper the size of a fingernail, which it will lift off the desk.

Those few minutes with a roll of tape contain most of Topic 10.2 of the AP course. There must be two kinds of the thing the tape acquired, because some pairs repel and others attract. Like kinds repel and unlike kinds attract. A charged object attracts neutral ones too, such as your finger and the scrap of paper, which needs its own explanation. And nothing was added to the tape that was not already in it: the tape and the table simply rearranged what they had. This lesson follows the tape through each of those observations.

What charge is

Electric charge is a fundamental property of matter, like mass, and it comes in two kinds, called positive and negative by convention. Every electron carries a charge of -e, every proton +e, and a neutron carries none. The magnitude e is the elementary charge, the smallest amount of charge found on any free particle:

e = 1.602 176 634 × 10-19 C

The value is exact, because since 2019 the coulomb (C) has been defined by fixing it, and the exam's table rounds it to 1.60 × 10^-19 C. The coulomb is a large unit: it takes 1/e = 6.24 × 10¹⁸ electrons to make -1 C. Any object's net charge is a whole number of elementary charges, because it is the difference between the numbers of protons and electrons it contains. An ordinary object has huge and almost exactly equal numbers of both, which is why it is neutral.

When the size and shape of a charged object do not matter for the question being asked, as for two small charged spheres far apart compared with their size, physicists model it as a point charge, all its charge at a single point. The next lesson uses that model to calculate forces.

Where the tape's charge came from: a transfer, never a creation

Peeling the stacked strips apart pulled electrons from one strip to the other. One strip ended with extra electrons and a net negative charge; the other was left short of exactly the same number, with a net positive charge of the same size. Which strip gets which sign depends on the brands of tape, but the equality does not. The same happens whenever two different materials are rubbed or pressed together and separated: OpenStax gives the classic case of glass rubbed with silk, where the glass becomes positive and the silk negative because electrons move from the glass to the silk. In solids it is always electrons that move. The protons are locked in nuclei and stay put.

That is the conservation of charge: the net charge of a system changes only if charge is transferred into or out of it, and the total charge in any process is constant. The tape-and-table system had zero net charge before peeling and has zero after; the charge was separated, not made.

How many electrons move? Suppose a strip ends with a charge of -2.4 × 10^-8 C. The number of extra electrons is N = |q|/e = (2.4 × 10-8 C)/(1.60 × 10-19 C) = 1.5 × 1011. That sounds like a lot until you compare it with what is there. A tenth of a gram of plastic tape contains roughly 3 × 10²² electrons, about one for every two nucleons, so the peel moved about five electrons in every million million. As OpenStax puts it, only a tiny fraction of the charges are involved.

Givens. Two identical metal spheres on insulating stands carry +6.0 nC and -2.0 nC. They are touched together and then separated.

The set-up in words. Two small polished spheres, each on a plastic rod, are pushed into contact for a moment and pulled apart again, without anything else touching them.

Principle. Conservation of charge for the two-sphere system, and the fact that charge moves freely through metal, so identical spheres in contact share the total equally.

Algebra. The total is qtotal = (+6.0 nC) + (-2.0 nC) = +4.0 nC, so each sphere ends with +4.0 nC/2 = +2.0 nC. Electrons flowed from the negative sphere to the positive one: the sphere that was +6.0 nC gained 4.0 nC worth of electrons, 2.5 × 10¹⁰ of them, and the sphere that was -2.0 nC lost the same number.

Sanity check. The final total, 2.0 nC + 2.0 nC, equals the initial total, 4.0 nC. Each sphere's charge moved toward the average, as a sharing process should.

The variant that flips the result. Start instead with +6.0 nC and -10.0 nC. The total is now -4.0 nC, and each sphere ends at -2.0 nC. The sphere that began strongly positive ends negative: the sign of the outcome is set by the sign of the total, not by which sphere started larger.

The core of it: charging never creates charge. It moves electrons from one place to another, and the books always balance.

Why a charged strip lifts a neutral scrap of paper

The scrap of paper has no net charge, so why is it pulled at all? Because a neutral object is still full of charges, and a nearby charged object can rearrange them. Bring the negative strip close: inside each molecule of the paper, the electron cloud is pushed slightly away from the strip and the positive nuclei are pulled slightly toward it. The paper stays neutral overall, but its near side is now slightly positive and its far side slightly negative. This is induced charge separation, or polarization, and it can happen in any neutral object.

Now compare the two forces on the paper. The strip attracts the paper's near, positive side and repels its far, negative side. Those induced charges are equal in size, but the attracted side is closer, and the electric force weakens rapidly with distance, as the next lesson quantifies. The attraction wins, and the paper jumps up. Reverse the strip's charge and the induced charges reverse too, so the paper is attracted again. A neutral object is attracted by a charge of either sign, which is why a charged strip attracts your finger whatever its sign, and why only repulsion is a reliable sign that two objects are both charged.

The same effect explains a rubbed balloon sticking to a wall and a charged comb bending a thin stream of tap water. Water molecules have a built-in separation of charge, so they swing round to point their opposite end at the comb; dissolved ions in tap water add to the effect. The stream curves toward the comb whichever sign the comb has.

Conductors, insulators, and the ground

In a conductor, such as a metal, some electrons are not tied to any one atom and move freely through the material, so charge placed on a conductor spreads out over it at once, and a nearby charge can push those electrons from one end of the object to the other. Salty water conducts too, through its free ions. In an insulator, such as glass, plastic or dry paper, the electrons are bound to their atoms. Charge put on an insulator stays where it was put, which is why you can charge a strip of tape at all, and polarization in an insulator is only the slight shift of charge within each molecule.

Whether a material conducts well or polarizes easily is a property of how its atoms are arranged and how tightly they hold their electrons. Physicists summarise how strongly a material polarizes in an electric field by its electric permittivity; the next lesson meets the permittivity of empty space, ε₀, inside Coulomb's law.

Grounding means connecting a charged object electrically to something so large that it can give or take electrons without its own charge changing noticeably, in practice the Earth. Touching a charged metal sphere with your finger grounds it through your body: if the sphere is negative, its extra electrons spread away into you and the floor; if it is positive, electrons flow up from the ground to neutralise it.

Charging by induction: the order of the steps decides the answer

Grounding makes it possible to charge a conductor without ever touching it with the charged object.

Givens. A neutral metal sphere stands on an insulating rod. You have a negatively charged plastic rod and your own finger.

The set-up in words. Four steps, in this order: bring the rod near the sphere without touching it; touch the far side of the sphere with your finger; take your finger away; take the rod away.

Principle. Like charges repel, electrons in a conductor move freely, and the ground can supply or absorb electrons; conservation of charge then fixes the result.

1. Rod near: the rod's negative charge pushes the sphere's free electrons to its far side. The sphere is polarized, still neutral overall.
2. Finger on: the repelled electrons now have a path to the ground and some of them leave through your body. The sphere now has fewer electrons than protons.
3. Finger off: the path is broken while the rod is still holding the remaining electrons away. The shortage is trapped.
4. Rod away: the remaining electrons spread evenly over the sphere, which is left with a net positive charge, opposite in sign to the rod.

Sanity check. The rod never touched the sphere and still has all its charge, so you can repeat the process as often as you like. The sphere's positive charge is exactly the charge of the electrons that left through your finger, so charge is conserved.

The variant that flips the result. Swap steps 3 and 4: take the rod away while your finger is still on the sphere. With nothing repelling them, electrons flow back up from the ground until the sphere is neutral again, and when you finally lift your finger it carries no charge at all. Same rod, same sphere, same finger, a different order, and the answer changes from positive to neutral. Charging by contact, for comparison, gives the sphere the same sign as the rod, because the rod hands some of its own charge over.

Every push you feel is electric

The forces you met in your first course as normal force, friction and tension are electric forces in disguise. When your hand presses on a table, the outer electrons of the atoms in your skin and in the wood repel each other strongly at very short range, and that repulsion, summed over an enormous number of atoms, is the normal force. Friction is the electric attraction and repulsion between atoms where two surfaces touch, and the tension in a string is the electric binding between its molecules. Treating these as separate contact forces is a convenience, because adding up the atomic interactions one by one would be impossible.

Common misconceptions

• "Rubbing creates electric charge." Rubbing separates charge. Electrons move from one material to the other, and the two end with equal and opposite charges.
• "A positively charged object has gained protons." In solids only electrons move. A positive object has lost electrons.
• "A neutral object contains no charge." It contains enormous, equal amounts of positive and negative charge, and those can be separated by a nearby charge.
• "Attraction proves two objects have opposite charges." A charged object also attracts a neutral one, by polarization. Only repulsion proves that both objects are charged, with the same sign.
• "Insulators cannot be charged." They charge readily; the charge simply cannot move through them, so it stays where it was placed.
• "Charging by induction gives the object the same sign as the rod." Induction with grounding gives the opposite sign. Contact gives the same sign.

Putting it together

Two strips of tape peeled the same way repel, a stacked pair pulled apart attract, and every charged strip lifts a neutral scrap of paper. Charge comes in positive and negative amounts that are whole multiples of e = 1.602 176 634 × 10^-19 C, exactly; electrons carry -e and protons +e. Charging moves electrons and never creates charge: a strip at -2.4 × 10^-8 C has 1.5 × 10¹¹ extra electrons, a tiny fraction of those present, and identical spheres at +6.0 nC and -2.0 nC share their +4.0 nC total to end at +2.0 nC each, while +6.0 nC and -10.0 nC end at -2.0 nC each. A charged object attracts a neutral one because it polarizes it and the nearer induced charge wins. Conductors let charge move and insulators hold it in place; grounding connects an object to the Earth's supply of electrons. Charging a sphere by induction with a negative rod leaves it positive if the ground connection is broken before the rod is removed, and neutral if the rod goes first. Underneath the everyday contact forces, it is all electric.

Sources

1. OpenStax. (2022). 18.1 Static electricity and charge: Conservation of charge. In College Physics 2e. openstax.org
2. OpenStax. (2022). 18.2 Conductors and insulators (charging by contact and by induction, polarization). In College Physics 2e. openstax.org
3. National Institute of Standards and Technology. (2024). CODATA 2022 value: elementary charge (exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
4. National Institute of Standards and Technology. (n.d.). SI redefinition: Ampere (from 20 May 2019 the ampere, and so the coulomb, is based on the fixed elementary charge). nist.gov
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 10.1 and 10.2. apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Balloons and static electricity (optional exploration of charging and polarization). phet.colorado.edu

Two forces between the same two particles

In the Bohr model of hydrogen, which Lesson 29 builds, the electron circles the proton at a distance of 5.29 × 10^-11 m. At that distance the proton pulls on the electron electrically with a force of 8.24 × 10^-8 N and gravitationally with a force of 3.63 × 10^-47 N. The electric force is larger by a factor of 2.27 × 10³⁹, the same ratio OpenStax finds in its own worked example. And yet it is gravity, not electricity, that holds the Moon in its orbit and you on the ground.

This lesson works through the law that produced the first of those numbers, step by step: first two charges, then three on a line, then three that are not in a line, where the forces have to be added as vectors. Each worked case ends with a variant in which one change reverses the answer. Then it explains why the weakest force in the pair ends up running the solar system.

Coulomb's law

The French physicist Charles Coulomb (1736-1806) was the first to propose a formula for the force between two charges. For two point charges q₁ and q₂ a distance r apart, the magnitude of the electric force each exerts on the other is

|FE| = k|q1q2|/r2 = (1/4πε0)|q1q2|/r2

The constant ε₀ is the permittivity of free space, which NIST's CODATA 2022 value gives as 8.854 187 8188 × 10^-12 C²/(N·m²), a measured quantity uncertain only in its last two digits. It makes the Coulomb constant k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C², which the exam's table rounds to 9.0 × 10⁹. Three features of the law matter as much as the formula.

• Direction. The force acts along the line joining the two charges. Like charges repel, so each force points away from the other charge; unlike charges attract, so each points toward it. Use magnitudes in the formula and decide the direction from the signs.
• Newton's third law. There is one force for the pair, of one size, acting on each charge in opposite directions. A charge of +4.0 μC and one of +1.0 μC push on each other equally hard.
• The inverse square. The force is proportional to each charge and to 1/r². It is an inverse-square law with exactly the form of Newton's law of gravitation from your first course, with charge in place of mass.

The proportionalities give quick answers without a calculator, which the exam rewards:

Step one: three charges on a line

Givens. A charge q₁ = +3.0 μC is fixed at x = 0 and q₂ = -2.0 μC at x = 0.30 m. A third charge, q₃ = +1.0 μC, sits at x = 0.60 m. Find the net force on q₃.

The set-up in words. Three small charged beads on a straight horizontal rod, the large positive one at the left end, the negative one in the middle, and the small positive one 0.30 m to the right of the middle one.

Principle. Coulomb's law for each pair acting on q₃, and the superposition principle: the net force is the vector sum of the separate forces. On a line, that means adding with signs for direction.

1. Force from q₁, 0.60 m away, repulsive, so toward +x: F13 = (8.99 × 109 N·m2/C2)(3.0 × 10-6 C)(1.0 × 10-6 C)/(0.60 m)2 = 0.075 N.
2. Force from q₂, 0.30 m away, attractive, so toward q₂, which is toward -x: F23 = (8.99 × 109)(2.0 × 10-6)(1.0 × 10-6)/(0.30)2 = 0.200 N.
3. Net force: Fnet = +0.075 N - 0.200 N = -0.125 N, that is, 0.125 N toward -x.

Sanity check. The middle charge is smaller than q₁ but half as far away, and halving the distance quadruples the force, so it should win: (2/3) × 4 = 2.67 times the pull of q₁, and 0.200/0.075 = 2.67. The units are N·m²/C² × C²/m², which is newtons.

The variant that flips the result. Move q₃ out to x = 2.00 m. Now the forces are (8.99 × 109)(3.0 × 10-12)/(2.00)2 = 6.7 × 10-3 N toward +x from q₁ and (8.99 × 109)(2.0 × 10-12)/(1.70)2 = 6.2 × 10-3 N toward -x from q₂. The net force, 5 × 10^-4 N, points toward +x: the direction has reversed. Far away, the two distances become nearly equal, and the larger charge wins. In between there is a point where the forces balance, where 3.0/x² = 2.0/(x - 0.30)²; taking square roots, √3(x - 0.30) = √2 x, so x = 1.63 m. Closer than that, q₃ is pulled toward the negative charge; farther, it is pushed away from the whole group.

What matters here: on a line, the nearer charge does not automatically win. Compare charge over distance squared for each one, and only then decide the direction.

Step two: forces that do not line up

Givens. A charge q = +1.0 μC sits at the origin. A charge of +4.0 μC is at (0, 0.30 m), directly above it, and a charge of -4.0 μC is at (0.40 m, 0), directly to its right. Find the magnitude and direction of the net force on q.

The set-up in words. Picture a right-angled triangle with q at the right-angled corner, the positive charge 0.30 m up the vertical side and the negative charge 0.40 m along the horizontal side.

Principle. Coulomb's law for each pair, then vector addition in components, because the two forces act along different lines.

1. From the +4.0 μC charge above: repulsion, pushing q straight down, along -y. F = (8.99 × 109)(4.0 × 10-6)(1.0 × 10-6)/(0.30)2 = 0.400 N.
2. From the -4.0 μC charge to the right: attraction, pulling q along +x. F = (8.99 × 109)(4.0 × 10-12)/(0.40)2 = 0.225 N.
3. Components of the net force: F_x = +0.225 N, F_y = -0.400 N. Magnitude: |F| = √(0.2252 + 0.4002) N = √0.210 N = 0.459 N. Direction: θ = tan-1(0.400/0.225) = 61° below the +x axis.

Sanity check. The magnitude, 0.459 N, is larger than either force alone but less than their sum, 0.625 N, as it must be for two perpendicular forces. The net force leans more toward -y than toward +x because the charge above is closer.

The variant that flips the result. Change the charge on the right from -4.0 μC to +4.0 μC. Its force on q is now a repulsion of the same size, 0.225 N, but along -x. The net force has the same magnitude, 0.459 N, and now points 61° below the -x axis: the horizontal component has reversed and the force has swung from the lower right to the lower left. Adding the magnitudes, 0.400 N + 0.225 N, would have given the same wrong answer, 0.625 N, in both cases, which is the clearest sign that forces must be added as vectors.

Why gravity usually loses, and why it rules the sky anyway

For two protons, which have both charge and mass, the ratio of the electric to the gravitational force is

FE/FG = ke2/(Gmp2) = (8.99 × 109)(1.60 × 10-19)2/[(6.67 × 10-11)(1.67 × 10-27)2] = 1.24 × 1036

using the CODATA values of the gravitational constant and the proton mass. Both forces fall off as 1/r², so the distance cancels and the ratio is the same at any separation. For any two objects that carry both charge and mass, the gravitational force is usually tiny compared with the electric one.

So why does gravity govern the planets? Because electric forces come in two signs and gravity in one. Large bodies contain almost exactly equal amounts of positive and negative charge, and their electric pushes and pulls cancel, while every kilogram attracts every other kilogram and nothing cancels gravity. A quick estimate shows how exact the cancellation has to be. NASA lists the masses of the Earth and the Moon as 5.97 × 10²⁴ kg and 7.3 × 10²² kg. A charge of the same sign on each, large enough to cancel their gravitational attraction, satisfies kQ² = GM_EM_M, so

Q = √(GMEMM/k) = √[(6.67 × 10-11)(5.97 × 1024)(7.3 × 1022)/(8.99 × 109)] C = 5.7 × 1013 C

That is about 3.6 × 10³² extra electrons, which sounds huge until you compare it with the roughly 1.8 × 10⁵¹ electrons the Earth contains, about one for every two nucleons. An imbalance of one electron in every five billion billion would undo the Moon's orbit. Matter in bulk is neutral far more precisely than that, so gravity, weak but uncancelled, wins at the scale of planets.

What matters here: electric forces dominate between individual particles; gravity dominates between large neutral bodies. The College Board phrases it the same way: gravitational forces dominate at large scales because systems at large scales tend to be electrically neutral.

Permittivity: what matter does to the electric force

The constant ε₀ describes empty space. Inside a material, the charges of each molecule shift in response to a nearby charge, as the last lesson described, and the shifted charges partly cancel that charge's field. Physicists say a material has its own permittivity, different from ε₀, and express the difference by a dielectric constant κ, the factor by which the material reduces the field of a given charge. A material whose electrons rearrange easily has a large κ. Air has κ = 1.00059, so for all practical purposes air behaves like empty space and Coulomb's law applies unchanged. Water has κ of about 80. Two ions dissolved in water therefore attract each other about eighty times more weakly than they would in a vacuum at the same distance. The same reduction is put to use in capacitors, the subject of Lesson 10.

Common misconceptions

• "The larger charge exerts the larger force." The force is mutual and has one size. A +4.0 μC charge and a +1.0 μC charge 0.30 m apart push on each other with 0.40 N each.
• "Doubling the distance halves the force." It divides the force by four: the law is inverse-square.
• "The nearest charge always decides the direction." In the line example the nearer -2.0 μC charge wins at x = 0.60 m but loses at x = 2.00 m, because what matters is each charge divided by its distance squared.
• "Forces from several charges add up as plain numbers." They add as vectors. Two perpendicular forces of 0.400 N and 0.225 N combine to 0.459 N, not 0.625 N.
• "Gravity is the stronger force, because it is the one we feel." Between two protons the electric force is 10³⁶ times stronger. We feel gravity because the Earth's charges cancel and its mass does not.

Pulling it together

Inside a hydrogen atom the electric attraction is 2.27 × 10³⁹ times the gravitational one. Coulomb's law gives the size of the force between point charges as k|q₁q₂|/r², with k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C² and ε₀ = 8.854 187 8188 × 10^-12 C²/(N·m²); the force acts along the line between the charges, repels like and attracts unlike charges, has one size for both members of a pair, and falls to a quarter when the distance doubles. With several charges, add the forces as vectors: on a line a +1.0 μC charge feels 0.125 N toward the -2.0 μC charge at x = 0.60 m but is pushed outward at x = 2.00 m, the balance point being 1.63 m; at the corner of a triangle the forces of 0.400 N and 0.225 N add in components to 0.459 N at 61°. Between two protons the electric force exceeds gravity by 1.24 × 10³⁶ at any distance, yet gravity rules the Moon's orbit because bulk matter is neutral to better than one part in 10¹⁸. In a material, polarization reduces the electric force by the dielectric constant, about 80 for water and essentially 1 for air.

Sources

1. OpenStax. (2022). 18.3 Coulomb's law. In College Physics 2e (Example 18.1: the electric and gravitational forces in hydrogen). openstax.org
2. National Institute of Standards and Technology. (2024). CODATA 2022 values: vacuum electric permittivity; Newtonian constant of gravitation; proton mass; electron mass; Bohr radius. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
3. OpenStax. (2022). 19.5 Capacitors and dielectrics (Table 19.1, dielectric constants of air and water). In College Physics 2e. openstax.org
4. Nave, R. (n.d.). Dielectrics. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
5. NASA Space Science Data Coordinated Archive. (n.d.). Planetary fact sheet: Metric (masses of the Earth and the Moon). NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 10.1, Electric charge and electric force (calculations limited to four or fewer charges). apcentral.collegeboard.org

The explanation everyone repeats

The usual explanation of why a car is a safe place in a thunderstorm is that its rubber tyres insulate it from the ground. The forecast office of the US National Weather Service in Fort Worth puts its own position bluntly: rubber-soled shoes and rubber tires provide no protection from lightning. What does protect you, it says, is the steel frame of a hard-topped vehicle, as long as you are not touching metal.

The tyre explanation is worth taking apart, because it fails in an instructive way. Electricity is known to be stopped by rubber; tyres are rubber; therefore tyres stop lightning. Each step sounds reasonable, and the conclusion is wrong. This lesson builds the idea needed to see why, the electric field, and then comes back to trace exactly where the reasoning broke.

First check: the scale of the claim

Before any new physics, test the claim against what lightning has already done by the time it reaches the car. It has just crossed a long column of air, and air is an excellent insulator: OpenStax's table gives its dielectric strength, the field it can stand before it breaks down and conducts, as 3 × 10⁶ V/m. A discharge that has forced its way through that much air will not be stopped by the last few centimetres of rubber. And there is a second problem: if the tyres really did insulate the car from the ground, a strike would leave the car's metal charged with nowhere for the charge to go, which is the opposite of safe. Whatever protects the occupants, it is not an insulating layer underneath them. To find out what it is, we need a way to describe what charges do to the space around them.

What an electric field is

A charge exerts a force on any other charge placed near it. Rather than recompute Coulomb's law for every possible second charge, physicists describe the space around the first charge by the force it would exert on a unit of charge placed at each point. The electric field at a point is the electric force on a small positive test charge placed there, divided by that charge:

E = FE/q

Its unit is the newton per coulomb, N/C. A test charge is chosen small enough that it does not disturb the charges producing the field, so the field belongs to the place, not to the test charge. Once the field is known, the force on any charge q placed there is F = qE: along the field for a positive charge, opposite to the field for a negative one.

For a single point charge q, Coulomb's law divided by the test charge gives the field at distance r:

|E| = k|q|/r2

The field points away from a positive charge and toward a negative one. The electric field is a vector, so the field of several charges at a point is the vector sum of their separate fields, the same superposition used for forces in the previous lesson.

Worked problem: the field between two charges

Givens. A charge of +2.0 nC sits at x = 0 and a charge of -2.0 nC at x = 0.20 m. Find the field at the midpoint, x = 0.10 m, and the force on an electron placed there.

The set-up in words. Two small charged beads 20 cm apart on a horizontal line, positive on the left, negative on the right, and an empty point exactly between them.

Principle. The field of each point charge, k|q|/r², with its direction set by the sign of the charge; then vector addition; then F = qE.

1. Field of each charge at the midpoint, 0.10 m from each: E = (8.99 × 109 N·m2/C2)(2.0 × 10-9 C)/(0.10 m)2 = 1.8 × 103 N/C.
2. Directions: the positive charge's field points away from it, toward +x; the negative charge's field points toward it, also toward +x. They add: Enet = 3.6 × 103 N/C toward +x, toward the negative charge.
3. Force on an electron: |F| = |q|E = (1.60 × 10-19 C)(3.6 × 103 N/C) = 5.8 × 10-16 N, directed toward -x, opposite to the field, toward the positive charge. Its acceleration would be F/m = (5.8 × 10^-16 N)/(9.11 × 10^-31 kg) = 6.3 × 10¹⁴ m/s².

Sanity check. N·m²/C² times C divided by m² is N/C. The electron is pulled toward the positive charge and pushed from the negative one, so the force must point toward -x, and it does. The enormous acceleration is typical of electrons, whose mass is tiny.

The variant that flips the result. Change the right-hand charge to +2.0 nC. Each field still has size 1.8 × 10³ N/C at the midpoint, but now each points away from its own positive charge, so they point in opposite directions and cancel. The net field at the midpoint is zero, and an electron placed there feels no force at all. One change of sign takes the answer from 3.6 × 10³ N/C to nothing.

Off the line, the same rules apply with components. At a point 0.10 m directly above the midpoint of the original +2.0 nC and -2.0 nC pair, each charge is √(0.10² + 0.10²) = 0.141 m away, so each field has size (8.99 × 10⁹)(2.0 × 10^-9)/(0.141)² = 9.0 × 10² N/C, at 45° to the horizontal. The positive charge's field points up and to the right, the negative charge's field down and to the right. The vertical components cancel and the horizontal ones add, giving 2 × 900 × cos 45° = 1.3 × 10³ N/C, parallel to the line of the charges, pointing from the positive side toward the negative side.

The upshot: find each field's size from k|q|/r², give it a direction from the sign of its charge, and add the vectors. The field at a point does not depend on what, if anything, is placed there.

Pictures of a field: vector maps and field lines

A field vector map draws an arrow at each of many points, pointing along the field there, with its length showing the field's strength. For a single positive charge the arrows point radially outward and shrink rapidly with distance: at twice the distance an arrow is a quarter as long. A field-line diagram is a simplified version that joins the arrows into continuous lines. Its rules are worth learning as a list, because exam questions ask you to draw and read these diagrams.

• Lines begin on positive charges and end on negative charges, or run off to infinity.
• The field at any point is tangent to the line through it, pointing along the line's direction.
• Lines are crowded together where the field is strong and spread apart where it is weak, and the number of lines leaving a charge is proportional to its charge.
• Lines never cross, because the field has only one direction at each point.

Three pictures cover most questions. Around a single positive charge, lines radiate outward like spokes; around a negative one they point inward. For a positive and a negative charge of equal size, a pair called a dipole, lines leave the positive charge, curve out and around, and arrive at the negative charge, with the straightest and densest lines running directly between them. For two equal positive charges, lines leave both and bend away from each other, and directly between the charges is a point with no field at all, the cancellation found in the worked variant above. A field line is not the path a charge must follow: it gives the direction of the force, and so of the acceleration, not of the velocity.

Inside a conductor, the field is zero

Now put charge on a conductor and wait a moment, until nothing is moving any more: the conductor has reached electrostatic equilibrium. In that state the electric field everywhere inside the metal is zero. The argument is short. A conductor is full of free electrons. If there were any field inside, it would push on them and they would move, so the conductor would not be in equilibrium. The electrons rearrange themselves until their own field exactly cancels any field inside, which takes far less than a second.

Three consequences follow, and OpenStax lists them as the properties of a conductor in equilibrium. Excess charge resides entirely on the outer surface, because like charges repel and spread as far apart as they can. Just outside the surface, the field is perpendicular to it, since any component along the surface would push charges sideways. And for a conducting sphere with charge Q, the charge spreads evenly over the surface, so outside the sphere the field is exactly that of a point charge Q at the centre.

Take a metal sphere of radius 0.10 m carrying +5.0 nC. Anywhere inside the metal, E = 0. Just outside the surface, E = (8.99 × 10⁹)(5.0 × 10^-9)/(0.10)² = 4.5 × 10³ N/C, pointing straight outward. At 0.30 m from the centre, E = (8.99 × 10⁹)(5.0 × 10^-9)/(0.30)² = 5.0 × 10² N/C, a ninth of the surface value at three times the distance, just as for a point charge.

Charge on an insulator behaves differently. It cannot move freely, so excess charge stays near where it was placed and can sit in the interior as well as on the surface, and the field inside an insulator need not be zero. The exam asks only for a qualitative description of fields inside insulators.

Tracing the error back

Return to the car. Its metal body is a closed conducting shell, a version of what is called a Faraday cage. Charge that arrives on a conductor stays on its outer surface, and the field inside a closed conducting shell is essentially zero. A lightning strike is not a static situation, since a large current flows, but the same behaviour carries over: the current runs over the outer metal of the car and on to the ground, around the people inside rather than through them. OpenStax describes exactly this case: the car's metal body works like a Faraday cage, and the inside is unaffected provided you stay completely inside.

So the tyre explanation failed at its first step. It looked for protection in an insulator under the car, when the protection comes from a conductor around it. It also assumed the danger was charge getting into the car, when the real point is that charge on a conductor goes to the outside and stays there. The same idea answers another of the College Board's questions for this unit, how to protect electronics from a burst of electromagnetic radiation: enclose them in a continuous metal box.

Common misconceptions

• "Rubber tyres protect a car from lightning." The National Weather Service says they provide no protection. The metal shell protects the occupants, because charge stays on a conductor's outer surface.
• "The field at a point depends on the test charge placed there." The field is set by the source charges. The test charge only reveals it, which is why it must be small.
• "Field lines are the paths charges travel along." They show the direction of the force at each point. A moving charge's velocity can point in quite a different direction.
• "A negative charge's field points away from it." Field direction is defined by the force on a positive test charge, so the field of a negative charge points toward it.
• "Charge on a conductor spreads through its whole volume." In equilibrium it sits entirely on the surface, and the field inside the metal is zero.

What you now know

Rubber tyres do not protect a car from lightning; its metal shell does, and seeing why takes the electric field. The field is the force per unit test charge, E = F/q, measured in N/C; a point charge makes a field of size k|q|/r², pointing away from it if positive and toward it if negative, and fields from several charges add as vectors. At the midpoint of +2.0 nC and -2.0 nC charges 0.20 m apart the field is 3.6 × 10³ N/C toward the negative charge, and an electron there is pushed the other way with 5.8 × 10^-16 N; make both charges positive and the midpoint field is zero. Vector maps and field lines picture the field: lines run from positive to negative charge, crowd where it is strong, and never cross. In electrostatic equilibrium the field inside a conductor is zero, excess charge lives on its outer surface, and the field just outside is perpendicular to it; outside a charged sphere the field is that of a point charge at the centre, 5.0 × 10² N/C at 0.30 m from +5.0 nC. Insulators hold charge where it is put, inside and out. A closed metal shell, a Faraday cage, keeps its interior nearly field-free, which is why the people in a hard-topped car are safe as long as they stay inside.

Sources

1. National Weather Service, Fort Worth/Dallas Weather Forecast Office. (n.d.). Lightning: Myths vs. facts. weather.gov
2. OpenStax. (2022). 18.4 Electric field: Concept of a field revisited; 18.5 Electric field lines: Multiple charges. In College Physics 2e. openstax.org
3. OpenStax. (2022). 18.7 Conductors and electric fields in static equilibrium (Faraday cages; a car struck by lightning). In College Physics 2e. openstax.org
4. OpenStax. (2022). 19.5 Capacitors and dielectrics (Table 19.1, dielectric strength of air). In College Physics 2e. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 10.3, Electric fields, and the Unit 10 essential questions. apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Charges and fields (optional: place charges and read the field). phet.colorado.edu

A head-on shot at a gold nucleus

In 1909, Ernest Rutherford's group fired alpha particles with energies of about 5 MeV at a thin gold foil. Most went straight through. A few were deflected through large angles, and about one in 8000 came straight back toward the source, as if it had bounced off something small and hard. Rutherford's analysis, published in 1911, concluded that almost all of an atom's mass and all of its positive charge are packed into a nucleus. The nucleus is about 10^-15 m across, inside an atom about 10^-10 m across.

Here is the question to answer by the end of this lesson's first section: an alpha particle with 5.0 MeV of kinetic energy heads straight at a gold nucleus and comes back. How close did it get before it turned round? Newton's laws alone make this hard, because the repulsive force grows without limit as the distance shrinks. The acceleration changes continuously, so none of the constant-acceleration equations apply. Energy makes it easy, once we have an expression for the energy stored in the arrangement of two charges.

Electric potential energy of two charges

The electric potential energy of two point charges is the work an external force must do to bring them from infinitely far apart to their present separation r, without giving them any kinetic energy:

UE = kq1q2/r

Unlike Coulomb's law, this formula keeps the signs of the charges, and the sign carries meaning. For two like charges U is positive: you had to push them together against their repulsion, and they will fly apart if released, turning U into kinetic energy. For unlike charges U is negative: they attract, you would have to do work to separate them, and the pair is bound. Zero is the energy when the charges are infinitely far apart, which is the convention the exam uses. Picture the graph of U against r for two positive charges. It is a curve that is nearly zero at large distances and rises ever more steeply as r shrinks, a hill an approaching charge must climb. For a positive and a negative charge the curve is the same shape turned upside down, a well.

For example, +3.0 μC and -2.0 μC at 0.30 m have U = (8.99 × 109)(3.0 × 10-6)(-2.0 × 10-6)/(0.30) = -0.18 J: separating them completely would cost 0.18 J.

Worked problem: the distance of closest approach

Givens. An alpha particle (charge +2e) with 5.0 MeV of kinetic energy, far from a gold nucleus (charge +79e), heads straight at it. Treat the gold nucleus, 49 times heavier than the alpha particle, as fixed. Useful values: 1 MeV = 1.60 × 10^-13 J and e = 1.60 × 10^-19 C.

The set-up in words. The alpha particle approaches along a straight line through the centre of the nucleus and slows as it climbs the potential-energy hill. It stops for an instant at its closest point and is pushed back the way it came.

Principle. Conservation of energy. Far away, U is essentially zero and all the energy is kinetic. At the turning point the alpha particle is momentarily at rest, so all of it has become potential energy: K_initial = kq₁q₂/r_min.

1. Kinetic energy in joules: K = (5.0 MeV)(1.60 × 10-13 J/MeV) = 8.0 × 10-13 J.
2. Rearrange for the distance: rmin = kq1q2/K = (8.99 × 109 N·m2/C2)(2 × 1.60 × 10-19 C)(79 × 1.60 × 10-19 C)/(8.0 × 10-13 J).
3. Evaluate: the numerator is (8.99 × 10⁹)(3.20 × 10^-19)(1.264 × 10^-17) = 3.64 × 10^-26 J·m, so rmin = 4.5 × 10-14 m, or 45 femtometres (1 fm = 10^-15 m).

Sanity check. The units are N·m²/C² times C² divided by J, which is N·m²/J = m. The answer is about two thousand times smaller than an atom, and a few times larger than the gold nucleus itself. OpenStax's rule r = (1.2 fm)A^1/3 puts the gold nucleus's radius at (1.2 fm)(197)^1/3 = 7.0 fm. The alpha particle turns round well outside the nucleus, so a purely electric calculation is the right one. Rutherford's conclusion that the nucleus is tiny follows: the positive charge must be concentrated within 45 fm for a head-on alpha particle to be turned back at all.

The variant that flips the result. Double the energy to 10 MeV and aim at aluminium instead (charge +13e, mass number 27). Now rmin = (8.99 × 109)(3.20 × 10-19)(13 × 1.60 × 10-19)/(1.60 × 10-12 J) = 3.7 × 10-15 m, 3.7 fm. From the same rule, the radii of the two nuclei are 3.6 fm for aluminium and 1.9 fm for the alpha particle. So they would overlap at 5.5 fm, before the electric turning point is reached. (Aluminium is only seven times heavier than the alpha particle, so its recoil should not be ignored. Allowing for it raises the turning point to about 4.3 fm, still inside 5.5 fm.) The same calculation that described a clean electric bounce off gold now predicts its own breakdown: the particles touch, and the strong nuclear force of Lesson 31 takes over. Low charge and high energy flip the outcome.

Bottom line: when a force changes with distance, use energy. Write the kinetic energy at one point and the potential energy kq₁q₂/r at another, and conservation does the rest.

Potential: energy per unit charge

Potential energy belongs to a pair of charges. As with force and field, it is useful to strip out the test charge and describe the place instead. The electric potential V at a point is the electric potential energy per unit charge that a test charge would have there, V = U/q, measured in volts, where 1 V = 1 J/C. For a single point charge q, with V = 0 infinitely far away,

V = kq/r

and for several point charges the potentials simply add as numbers, with their signs: V = kΣqi/ri. No components, no angles. That is the great convenience of potential: it is a scalar.

What a charge feels is a potential difference. When a charge q moves between two points whose potentials differ by ΔV, its potential energy changes by

ΔUE = qΔV

and if no other force does work, its kinetic energy changes by the same amount in the opposite direction. This gives atomic physics its favourite unit of energy, the electron volt: the energy an electron gains falling through a potential difference of 1 V. Exactly, 1 eV = 1.602 176 634 × 10^-19 J. The 5.0 MeV alpha particle above had five million of them. Potential differences do not only come from fixed charges. In a battery, chemical reactions separate positive and negative charges and keep the terminals at a steady potential difference. So a 1.5 V cell gives each coulomb that passes through the circuit 1.5 J.

A short example. An electron starts from rest in an evacuated tube and is accelerated through a potential difference of 2500 V. It gains K = eΔV = (1.60 × 10-19 C)(2500 V) = 4.0 × 10-16 J, which is 2500 eV. Its speed becomes v = √(2K/m) = √[2(4.0 × 10-16 J)/(9.11 × 10-31 kg)] = 3.0 × 107 m/s, a tenth of the speed of light. Now send that 2500 eV electron toward a plate held 2000 V more negative than its starting point. It is moving against the field, so it slows down, but it arrives with 2500 - 2000 = 500 eV. Make the plate 3000 V more negative and the result flips. The electron runs out of energy after 2500 V of the climb, stops, and turns back without reaching the plate. Lesson 27 uses exactly this stopping trick to measure the energies of electrons knocked out of metals by light.

Potential and field are different things: one pair, two answers

Go back to the +2.0 nC and -2.0 nC charges 0.20 m apart from the previous lesson. At the midpoint, each contributes a potential of (8.99 × 10⁹)(2.0 × 10^-9)/(0.10) = 180 V, positive from one and negative from the other, so V = 180 V - 180 V = 0. But the field there, found last lesson, is 3.6 × 10³ N/C. Zero potential does not mean zero field. Now make both charges +2.0 nC. The potential at the midpoint becomes 180 V + 180 V = 360 V, while the two fields cancel and E = 0. Zero field does not mean zero potential either. The potential adds as a scalar; the field adds as a vector; and the field depends on how the potential changes from place to place, not on its value at one place.

Equipotentials and the slope of the potential

An equipotential is a line, or in three dimensions a surface, on which every point has the same potential. Equipotentials work like the contour lines on a hiking map: contours join points of equal height, and where they bunch together the slope is steep. In the same way, the electric field is perpendicular to the equipotentials everywhere, points in the direction in which the potential decreases, and is strongest where the equipotentials are closest. Its average magnitude between two points is the potential difference divided by the distance:

|E| = |ΔV/Δr|

This also shows that the volt per metre is the same unit as the newton per coulomb. Between two parallel metal plates 0.020 m apart held at a potential difference of 12 V, the field is (12 V)/(0.020 m) = 600 V/m, pointing from the higher-potential plate toward the lower one.

Three equipotential maps are worth being able to sketch from a description. Around a single positive point charge, the equipotentials are circles centred on the charge. Drawn for equal steps of potential, they crowd together near the charge and spread out far from it, because V = kq/r changes fastest at small r. Between a positive and a negative charge of equal size, each charge is surrounded by its own family of closed curves. The line exactly midway between them is the equipotential V = 0. Between two large parallel plates, the equipotentials are straight lines parallel to the plates and evenly spaced, which is the signature of a uniform field. Because the field has no component along an equipotential, moving a charge along one takes no work at all; work is done only in moving across them.

Bottom line: the field points downhill in potential, at right angles to the equipotentials, and its size is how steeply the potential falls with distance.

The energy of a group of charges

The total electric potential energy of several point charges is the sum of the potential energies of every pair. Place +1.0 nC, +1.0 nC and -1.0 nC at the corners of an equilateral triangle of side 0.10 m. Each pair contributes (8.99 × 10⁹)(1.0 × 10^-9)²/(0.10) = 9.0 × 10^-8 J in size: the two positive charges repel (+), and each of them attracts the negative charge (-, twice). The total is +9.0 - 9.0 - 9.0 = -9.0 × 10^-8 J, so the group is bound: it would take 9.0 × 10^-8 J of work to pull all three charges far apart. Make all three charges positive and every pair repels, giving +2.7 × 10^-7 J: the same triangle now stores energy that would fling the charges apart if they were released.

Conductors in contact share one potential

Connect two charged metal objects with a wire and electrons flow between them until their surfaces are at the same potential, not until they hold the same charge. For two spheres of different sizes joined this way, the larger sphere ends up holding more of the charge. For a sphere V = kq/R, and equal V with a larger R needs a larger q. This is why a conductor in electrostatic equilibrium is itself an equipotential, surface and interior alike, which fits with the zero field inside it found last lesson.

Common misconceptions

• "Zero potential means zero field." Midway between +2.0 nC and -2.0 nC the potential is zero and the field is 3.6 × 10³ N/C.
• "Zero field means zero potential." Midway between two +2.0 nC charges the field is zero and the potential is 360 V.
• "Potential and potential energy are the same thing." Potential is energy per unit charge, in volts, and belongs to a place; potential energy, in joules, belongs to a charge at that place, U = qV.
• "A negative potential energy means a mistake." It means the charges attract and the system is bound: energy must be added to separate them.
• "Charges always move toward lower potential." Positive charges released from rest do. Electrons, being negative, accelerate toward higher potential.
• "Moving a charge along an equipotential takes work." The field has no component along an equipotential, so no work is done.

The takeaway

A 5.0 MeV alpha particle fired head-on at a gold nucleus stops 4.5 × 10^-14 m from its centre, found by setting its kinetic energy equal to the potential energy kq₁q₂/r. At 10 MeV against aluminium the same arithmetic predicts a turning point inside the nuclei's combined radius, where the electric-only model fails. Electric potential energy, kq₁q₂/r with signs, is positive for repelling charges, negative for a bound pair, and adds over every pair in a group. Electric potential, V = kq/r per charge, is energy per unit charge and adds as a scalar. A charge crossing a potential difference changes its potential energy by qΔV. So an electron accelerated through 2500 V gains 2500 eV and reaches 3.0 × 10⁷ m/s, and a retarding difference larger than 2500 V stops it and turns it back.

Potential and field answer different questions: the midpoint of opposite charges has V = 0 and a strong field, the midpoint of like charges has E = 0 and V = 360 V. The field is perpendicular to the equipotentials, points toward lower potential, and has size |ΔV/Δr|, 600 V/m across 12 V and 2.0 cm. Conductors in contact come to the same potential, and a conductor in equilibrium is an equipotential throughout.

Sources

1. OpenStax. (2022). 19.1 Electric potential energy: Potential difference (the electron volt). In College Physics 2e. openstax.org
2. OpenStax. (2022). 19.3 Electrical potential due to a point charge; 19.4 Equipotential lines. In College Physics 2e. openstax.org
3. OpenStax. (2022). 30.2 Discovery of the parts of the atom: Electrons and nuclei (Rutherford's alpha-scattering results). In College Physics 2e. openstax.org
4. OpenStax. (2022). 31.3 Substructure of the nucleus (nuclear radius r = r₀A^1/3, r₀ = 1.2 fm). In College Physics 2e. openstax.org
5. National Institute of Standards and Technology. (2024). CODATA 2022 value: electron volt (exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 10.4, 10.5 and 10.7. apcentral.collegeboard.org

A drop of oil that will not fall

In 1910 Robert Millikan, working at the University of Chicago with his student Harvey Fletcher, published the first results of an experiment that would occupy him for years. An atomiser, like the one on a perfume bottle, sprayed a mist of oil droplets about a micrometre across. A few drifted through a pinhole into a chamber between two horizontal metal plates 16 mm apart. Millikan watched them through a telescope, timing each one against lines in the eyepiece. X-rays knocked electrons loose in the air, and now and then a droplet picked one up or lost one.

With the plates uncharged, a droplet fell slowly through the air, and its speed revealed its size and mass. With a high voltage across the plates, the electric force on its charge drove it upward, and its new speed revealed that force. Switching the voltage on and off, Millikan could follow one droplet up and down for a long time. Fletcher had found that a light oil sold for lubricating clocks gave drops that lasted.

Every charge Millikan measured was a whole-number multiple of one value. His 1913 paper in Physical Review, "On the elementary electrical charge and the Avogadro constant", gave that value within half a percent of the one accepted today. The two plates at the heart of the apparatus are a capacitor. This lesson follows the experiment's physics from them outward: how plates store charge and energy, why the field between them is uniform, and what that field does to a charged particle.

The parallel-plate capacitor

A parallel-plate capacitor is two flat, parallel conducting plates, separated by a gap and carrying charges of equal size and opposite sign, +Q and -Q. Connect the plates to a battery and electrons are pulled off one plate and pushed onto the other until the potential difference between the plates equals the battery's. The ratio of the charge on either plate to the potential difference between them is the capacitance:

C = Q/ΔV

Its unit is the farad, F, one coulomb per volt; real capacitors are measured in microfarads, nanofarads and picofarads. Capacitance depends only on the capacitor's physical construction, not on how much charge it happens to hold. For parallel plates of area A separated by a distance d, with a material of dielectric constant κ between them,

C = κε0A/d

with κ = 1 for a vacuum, and 1.00059, close enough to 1, for air. Bigger plates hold more charge at a given voltage, and so does a smaller gap, because the opposite charges then attract each other more strongly.

Givens. Two square metal plates 0.20 m on a side, 1.0 mm apart in air, connected to a 12 V battery.

The set-up in words. Two sheets the size of a large book cover, held flat and parallel with a gap about the thickness of a coin, one wired to each terminal of the battery.

Principle. C = ε₀A/d, then Q = CΔV, then the field from E = ΔV/d.

1. Area: A = (0.20 m)2 = 0.040 m2. Capacitance: C = (8.85 × 10-12 C2/(N·m2))(0.040 m2)/(1.0 × 10-3 m) = 3.5 × 10-10 F, or 354 pF.
2. Charge: Q = CΔV = (3.54 × 10-10 F)(12 V) = 4.2 × 10-9 C on each plate, +Q on one and -Q on the other.
3. Field: E = ΔV/d = (12 V)/(1.0 × 10-3 m) = 1.2 × 104 V/m, pointing from the positive plate to the negative one.

Sanity check. The exam's table gives a second route to the field, E = Q/(κε₀A) = (4.25 × 10^-9 C)/[(8.85 × 10^-12)(0.040 m²)] = 1.2 × 10⁴ V/m, the same answer. And a few hundred picofarads is what two book-sized sheets should give: a farad is enormous.

The variant that flips the result. Slide a sheet of paper, κ = 3.7, into the gap so that it fills it. What happens next depends on one thing: whether the battery is still connected. With the battery connected, ΔV stays at 12 V, so the field E = ΔV/d stays at 1.2 × 10⁴ V/m. Meanwhile the capacitance rises 3.7 times to 1.3 × 10^-9 F, and the battery pushes on 3.7 times the charge, 1.6 × 10^-8 C. Disconnect the battery first, and the charge is trapped at 4.2 × 10^-9 C. Now ΔV = Q/C falls 3.7 times to 3.2 V, and the field falls to 3.2 × 10³ V/m. The same sheet of paper leaves the field unchanged in one case and cuts it by a factor of 3.7 in the other.

Why this matters: in any capacitor question, first decide what is held fixed. A connected battery fixes ΔV; an isolated capacitor fixes Q. Everything else follows from C = Q/ΔV.

Why the field between the plates is uniform

Each plate, on its own, is a sheet of evenly spread charge. The field of a large charged sheet points straight away from it, or toward it if negative. It has the same strength at every distance, as long as you stay far from the edges. Between the plates, the field of the positive plate and the field of the negative plate point the same way, from + toward -, and add. Outside the plates they point in opposite directions and cancel.

So the field is confined to the gap and is the same in magnitude and direction everywhere in it. Drawn as field lines, it is a set of straight, parallel, evenly spaced lines running from one plate to the other. The equipotentials are evenly spaced lines parallel to the plates. Only near the edges do the lines bulge outward. The exam ignores that edge effect unless told otherwise.

A uniform field means a constant force, qE, on a charge anywhere between the plates. A constant force means the constant-acceleration mathematics of your first course applies, with qE/m in place of g.

Millikan's drop, balanced

Millikan worked from the rising and falling speeds of his drops. The simplest version of his measurement, used in classrooms and on exams, adjusts the voltage until a drop hangs perfectly still, so that the electric force exactly balances its weight. The physics is the same.

Givens. An oil drop of mass 4.9 × 10^-15 kg hangs motionless between horizontal plates 16.0 mm apart when the potential difference across them is 1600 V. The drop carries extra electrons.

The set-up in words. Two horizontal plates, one above the other, with a tiny drop floating in the uniform field between them.

Principle. Newton's first law: the upward electric force qE balances the downward weight mg; and E = ΔV/d between the plates.

1. Field: E = ΔV/d = (1600 V)/(0.0160 m) = 1.00 × 105 V/m.
2. Balance: |q|E = mg, so |q| = mg/E = (4.9 × 10-15 kg)(9.8 m/s2)/(1.00 × 105 V/m) = 4.8 × 10-19 C.
3. In elementary charges: (4.8 × 10-19 C)/(1.60 × 10-19 C) = 3.0. The drop carries three extra electrons.
4. Direction: the force on the negative drop must point up, and the force on a negative charge is opposite to the field, so the field points down. The top plate is the positive one.

Sanity check. V/m is N/C, so kg·m/s² divided by N/C gives coulombs. The answer comes out as a whole number of elementary charges, which is the whole point of the experiment; a result like 2.4e would mean a mistake somewhere.

The variant that flips the result. An X-ray frees an electron in the air nearby and the drop picks it up, so now it carries 4e. At 1600 V the electric force, 4e × E = 6.4 × 10^-14 N, now exceeds the weight, 4.8 × 10^-14 N, and the drop that was hanging still starts to rise. To hold it still again, the field must be three quarters of its old value, so the voltage must be lowered to 1200 V. A single electron decides whether the drop floats or climbs, which is how Millikan could see charge changing one electron at a time.

A charged particle crossing the plates: projectile motion again

Fire a charged particle into the gap parallel to the plates and it behaves like a ball thrown horizontally. It keeps its speed along the plates and accelerates steadily toward one of them.

An electron enters at 2.0 × 10⁷ m/s between plates 5.0 cm long with a field of 2.0 × 10⁴ V/m between them. It spends t = L/v = (0.050 m)/(2.0 × 107 m/s) = 2.5 × 10-9 s between the plates. Its acceleration toward the positive plate is a = eE/m = (1.60 × 10-19 C)(2.0 × 104 V/m)/(9.11 × 10-31 kg) = 3.5 × 1015 m/s2. So it is deflected sideways by y = (1/2)at2 = (1/2)(3.5 × 1015)(2.5 × 10-9)2 = 1.1 × 10-2 m, about a centimetre, along a parabola.

Gravity is utterly negligible: the electric acceleration is about 10¹⁴ times g. A proton at the same speed is pushed the other way, toward the negative plate, and because it is 1836 times heavier it is deflected only about 6 micrometres. Energy gives the same physics another way. As the electron drifts toward the positive plate it moves to higher potential, its potential energy qV falls, and its kinetic energy rises by exactly the same amount.

The energy stored in a capacitor

Separating charge takes work, and that work is stored as electric potential energy in the capacitor:

UC = (1/2)QΔV = (1/2)CΔV2 = Q2/(2C)

The factor of one half surprises people. While the capacitor charges, the potential difference across it rises steadily from zero to its final value. So the first bits of charge are moved across almost no potential difference, and the last across nearly all of it. On average, the charge crosses ΔV/2. The 354 pF capacitor at 12 V stores (1/2)(4.25 × 10^-9 C)(12 V) = 2.6 × 10^-8 J, a tiny amount. OpenStax's worked example of a heart defibrillator stores 400 J in an 8.00 μF capacitor charged to 10 000 V.

In the dielectric variant above, the connected capacitor's energy rises 3.7 times, because the battery supplies more charge at the same voltage. The isolated capacitor's energy falls 3.7 times, because its voltage drops with the charge unchanged. In the isolated case the missing energy went into pulling the paper into the gap.

What a dielectric does

A dielectric works by polarization, the same effect that lets a charged strip lift a scrap of paper. The field between the plates shifts the charges inside every molecule of the dielectric. That leaves a thin layer of negative charge on the face next to the positive plate, and positive charge next to the negative plate. Those induced layers produce a field inside the dielectric pointing opposite to the plates' field, so the net field is weaker for the same charge on the plates. A weaker field means a smaller potential difference for the same charge, and so a larger capacitance. Here are dielectric constants from OpenStax's table, at 20 °C:

A dielectric does one more useful thing: it can withstand a much stronger field than air before it breaks down and conducts, a limit called its dielectric strength. Air's dielectric strength is about 3 × 10⁶ V/m, so a 1.0 mm air gap sparks over at about (3 × 10⁶ V/m)(1.0 × 10^-3 m) = 3000 V.

Common misconceptions

• "A capacitor stores charge." Its net charge is zero: +Q on one plate and -Q on the other. It stores energy, in the separation of those charges.
• "Capacitance depends on the voltage or the charge." C = κε₀A/d depends only on the plates and the material between them. Doubling the voltage doubles the charge and leaves C unchanged.
• "The field is strongest right next to each plate." Away from the edges it is the same everywhere in the gap, which is what makes parallel plates so useful.
• "Inserting a dielectric always weakens the field." Only if the charge is fixed. With the battery connected, the field stays at ΔV/d and the charge increases instead.
• "The stored energy is QΔV." It is half of that, because the potential difference rose from zero while the capacitor charged.

Summing up

Millikan's drops, watched between horizontal plates 16 mm apart, always carried whole multiples of one charge, and his 1913 value of e was within half a percent of today's. Those plates were a parallel-plate capacitor, with equal and opposite charges. Its capacitance, C = Q/ΔV = κε₀A/d, depends only on construction: 354 pF for two 0.20 m plates 1.0 mm apart in air. Its uniform field, E = ΔV/d = Q/(κε₀A), is confined to the gap: 1.2 × 10⁴ V/m at 12 V. A drop of 4.9 × 10^-15 kg balanced at 1600 V across 16.0 mm carries 4.8 × 10^-19 C, exactly three electrons, with the top plate positive. One more electron and the voltage must fall to 1200 V, or the drop rises.

A charged particle crossing the plates follows a projectile's parabola with acceleration qE/m, a centimetre of deflection for a fast electron and micrometres for a proton. The stored energy is QΔV/2, half of QΔV because the voltage built up from zero. A dielectric, polarized by the field, multiplies the capacitance by κ. With a battery connected the field stays fixed and the charge and energy rise, while on an isolated capacitor the field, the voltage and the energy all fall.

Sources

1. Schirber, M. (2012). Landmarks: Millikan measures the electron's charge. Physics, 5, 9. American Physical Society. physics.aps.org
2. Millikan, R. A. (1913). On the elementary electrical charge and the Avogadro constant. Physical Review, 2(2), 109-143. https://doi.org/10.1103/PhysRev.2.109. Abstract page: journals.aps.org
3. OpenStax. (2022). 19.5 Capacitors and dielectrics (Table 19.1, dielectric constants and strengths). In College Physics 2e. openstax.org
4. OpenStax. (2022). 19.7 Energy stored in capacitors (Example 19.11, a defibrillator). In College Physics 2e. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 10.6 and 10.7 (parallel-plate capacitors only; edge effects ignored). apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Capacitor lab: Basics (optional: vary plate area and separation and watch C, Q and the field). phet.colorado.edu

A tenth of a millimetre per second

A copper wire 1.0 mm across carrying a current of 1.0 A is full of free electrons darting about at high speed in every direction. Their net drift along the wire, the motion that actually carries the current, is about 0.1 mm per second. Yet when you close a switch a lamp several metres away lights at once. The resolution is that the electrons do not have to travel from the switch to the lamp. The wire is full of them from the start, and closing the switch sets up an electric field along the whole circuit almost instantly. As OpenStax puts it, the high speed of electrical signals comes from the force between charges acting rapidly at a distance. Every electron in the circuit starts drifting at nearly the same moment.

This lesson builds the three quantities every circuit question uses: current, potential difference and resistance. It follows one procedure from start to finish, measuring how the current through an object depends on the potential difference across it. The object is first a resistor and then a light bulb, where one changed property bends the straight line.

Current: charge per second, with a direction but not a vector

Electric current is the rate at which charge passes through a cross-section of a conductor:

I = Δq/Δt

The unit, the ampere (A), is one coulomb per second. A phone charger delivering 2.0 A for 30 minutes moves Δq = IΔt = (2.0 A)(1800 s) = 3600 C through its cable, the charge of about 2.2 × 10²² electrons. Charge moves because of a potential difference, sometimes called an electromotive force or emf, ε, when it is supplied by a source such as a battery. If the current in part of a wire is zero, the net motion of the charges there is zero, although every individual electron is still moving quickly at random.

Current has a direction around a circuit, but it is not a vector. It does not point in a direction in space, only along the wire, and currents meeting at a junction add as plain numbers. By convention, the direction of conventional current is the direction positive charge would move: out of the positive terminal of a battery, through the external circuit and back into the negative terminal. In a metal the moving charges are actually electrons, drifting the opposite way. But a flow of negative charge one way is equivalent to a flow of positive charge the other, and every circuit rule works the same with either picture. The exam always uses conventional current.

The drift speed comes from counting carriers. Suppose a wire of cross-sectional area A has n free electrons per cubic metre, each of charge e, drifting at speed v_d. In one second the electrons in a length v_d of wire pass any point, so I = neAv_d. Copper has about 8.3 × 10²⁸ free electrons per cubic metre, one per atom. OpenStax uses this to find 0.45 mm/s for 20 A in a thick household wire, and for 1.0 A in a 1.0 mm wire the same count gives about 0.1 mm/s.

Circuits, and the symbols that describe them

A circuit is made of closed loops through which charge can flow, containing elements such as wires, batteries, resistors, bulbs, capacitors, switches, ammeters and voltmeters. A closed circuit is one in which charge can flow; an open circuit, such as one with a switch open or a broken wire, is one in which it cannot. A short circuit is a path through which charge can flow with no change in potential: a bare wire across a bulb's terminals is one, and it diverts the current around the bulb, which goes out. A single element can belong to more than one loop, which is what makes larger circuits interesting.

Circuits are drawn as schematic diagrams with standard symbols. Since this course cannot draw them, here they are in words, so that you recognise them on the exam. A battery is a pair of parallel lines of unequal length, the longer one marking the positive terminal. A resistor is a zigzag line. A bulb is a small circle with a loop or cross inside it. A switch is a gap in the line with a short hinged segment that can close it. A capacitor is two parallel lines of equal length. Ammeters and voltmeters are circles containing the letters A and V. A diagonal arrow drawn across any symbol marks that element as variable, such as a variable resistor.

Resistance, and what it depends on

The resistance R of an object measures how strongly it opposes the movement of charge through it: the potential difference needed per ampere of current, in ohms (Ω), where 1 Ω = 1 V/A. For a uniform wire or bar,

R = ρL/A

A longer conductor has more resistance, in proportion to its length L, because charge has further to go through the obstruction; a thicker one has less, in inverse proportion to its cross-sectional area A, because there are more parallel paths. The constant ρ is the resistivity, a fundamental property of the material that depends on its atomic structure, measured in Ω·m. Values from OpenStax's table at 20 °C:

The range is staggering: hard rubber resists at least 10²⁰ times as strongly as copper, which is why the same table covers both conductors and insulators. The resistivity of a metal typically increases with temperature, because hotter atoms vibrate more and scatter the drifting electrons more often.

Givens. A 10.0 m length of copper wire with a diameter of 0.50 mm, at 20 °C.

The set-up in words. A coil of thin copper wire such as you might use to wind an electromagnet, unrolled to its full length.

Principle. R = ρL/A, with the area of the circular cross-section A = πr².

1. Area: A = π(0.25 × 10-3 m)2 = 1.96 × 10-7 m2.
2. Resistance: R = (1.72 × 10-8 Ω·m)(10.0 m)/(1.96 × 10-7 m2) = 0.88 Ω.

Sanity check. Ω·m times m divided by m² is Ω. Less than an ohm for ten metres of wire is why connecting wires are normally treated as having no resistance; as the College Board notes, that is only safe when the circuit contains other elements whose resistance is much larger.

The variant that flips the result. Make the same wire from nichrome. Its resistivity is 58 times copper's, so R = (100 × 10^-8 Ω·m)(10.0 m)/(1.96 × 10^-7 m²) = 51 Ω. A wire you could ignore has become a component in its own right, the kind of wire used for heating elements. Halving the copper wire's diameter instead would quarter its area and quadruple its resistance to 3.5 Ω.

The procedure: measuring a resistor, then a bulb

Ohm's law states that for many conductors the current is proportional to the potential difference across them:

I = ΔV/R

with R constant. Materials and elements that obey it, whose resistance stays the same at every current, are called ohmic. The way to find out whether something is ohmic is to measure it. Connect the element in series with a variable source of potential difference, such as one to four cells in a holder, an ammeter in series to read the current, and a voltmeter connected across the element to read ΔV. For each setting, record ΔV and I, then plot current on the vertical axis against potential difference on the horizontal axis.

Givens. Readings for a resistor marked 220 Ω: 1.55 V and 7.0 mA; 3.10 V and 14.1 mA; 4.62 V and 21.0 mA; 6.15 V and 28.0 mA.

The set-up in words. A battery holder whose output can be tapped after one, two, three or four AA cells, a resistor, and a multimeter used first as an ammeter in series with the resistor and then as a voltmeter across it.

Principle. If the element is ohmic, I = (1/R)ΔV is a straight line through the origin, and its slope is 1/R.

1. The four points lie on a straight line through the origin, so the resistor is ohmic over this range.
2. Slope, from the origin to the last point: (28.0 × 10-3 A)/(6.15 V) = 4.55 × 10-3 A/V.
3. Resistance: R = 1/slope = 1/(4.55 × 10-3 A/V) = 220 Ω.

Sanity check. The answer matches the marked value, and each individual reading gives the same ratio: 1.55 V/7.0 mA = 221 Ω. Using a graph rather than a single reading averages out the scatter and would reveal at once if the line curved.

The variant that flips the result. Replace the resistor with a small torch bulb and repeat. Readings of the kind such a bulb gives: 0.10 V and 0.10 A, 0.50 V and 0.18 A, 1.00 V and 0.22 A, 1.50 V and 0.25 A, 2.00 V and 0.28 A, 2.50 V and 0.30 A. Now the graph is not a straight line: it rises steeply at first and then flattens. The ratio ΔV/I climbs from 1.0 Ω at the lowest reading to 8.3 Ω at 2.50 V. The bulb is non-ohmic. The reason is temperature: as the current rises, the tungsten filament heats until it glows, and tungsten's resistivity rises with temperature. OpenStax's worked example finds a filament of 0.350 Ω at room temperature rising to about 4.8 Ω at its working temperature of 2850 °C. The ohmic resistor stayed near room temperature, which is the real meaning of the College Board's statement that an ohmic material has a resistivity that does not change: it is ohmic as long as its temperature does not change much.

So what?: plot I against ΔV and read the slope as 1/R; a straight line through the origin means ohmic, a bending line means the resistance is changing, usually because the element is heating up. On the exam, resistors and bulbs are treated as ohmic unless a question says otherwise.

Where the energy goes

Charge passing through a resistor falls through a potential difference ΔV and loses electric potential energy qΔV. It does not speed up, because the drift speed is the same throughout a wire of uniform thickness, so the energy goes into the resistor's atoms, through the collisions that make up the resistance, and appears as thermal energy. Resistors therefore warm themselves and their surroundings, which is a nuisance in a computer and the whole point of a toaster or a kettle. The rate of that conversion, the power, is the subject of the next lesson.

Common misconceptions

• "Electrons race round the circuit at enormous speed." Their drift is around 0.1 mm/s. The lamp lights at once because the field is established along the whole circuit almost instantly.
• "A battery stores charge and pushes it out." The charges are already in every part of the circuit. The battery maintains a potential difference that gives each coulomb energy; a flat battery has run out of chemical energy, not of charge.
• "Current flows from the negative terminal, so conventional current is wrong." Electrons do drift from the negative terminal, but positive charge moving the other way is equivalent. Conventional current is a consistent convention, not a mistake.
• "A thicker wire has more resistance because it contains more metal." It has less: more cross-section means more parallel paths, and R = ρL/A.
• "Every conductor obeys Ohm's law." A bulb filament's resistance rises many times as it heats, so its I-V graph curves.

Looking back

Free electrons in a copper wire carrying 1.0 A drift at about 0.1 mm/s, and lamps still light at once because the field is set up along the whole circuit together. Current is the rate of flow of charge, I = Δq/Δt: 3600 C in half an hour at 2.0 A. It has a direction along the wire but is not a vector, and conventional current runs the way positive charge would move, opposite to the electrons in a metal. A circuit is closed, open or short-circuited, and its schematic uses standard symbols for cells, resistors, bulbs, switches, capacitors and meters. Resistance, R = ρL/A, grows with length and shrinks with cross-section: 0.88 Ω for 10.0 m of 0.50 mm copper, 51 Ω for the same wire in nichrome. Resistivity is a property of the material that rises with temperature in metals. Ohm's law, I = ΔV/R, holds for ohmic elements, whose I-V graph is a straight line through the origin with slope 1/R, as for a 220 Ω resistor. A bulb's graph curves because its hot filament's resistance climbs from 1 Ω to more than 8 Ω. The energy charges lose crossing a resistance becomes thermal energy in it.

Sources

1. OpenStax. (2022). 20.1 Current (drift velocity; speed of electrical signals). In College Physics 2e. openstax.org
2. OpenStax. (2022). 20.2 Ohm's law: Resistance and simple circuits. In College Physics 2e. openstax.org
3. OpenStax. (2022). 20.3 Resistance and resistivity (Tables 20.1 and 20.2; Examples 20.5 and 20.6 on a tungsten filament). In College Physics 2e. openstax.org
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 11.1, 11.2 and 11.3, and the exam conventions (resistors and bulbs ohmic unless stated). apcentral.collegeboard.org
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Resistance in a wire (optional: vary ρ, L and A). phet.colorado.edu

The first bulb gets the most

Connect two identical torch bulbs one after the other in a single loop with a 3.0 V battery, and ask which glows more brightly. A very common answer is the first one, the bulb the current reaches first after leaving the battery's positive terminal. It takes its share of the current, the reasoning goes, and leaves less for the second. The College Board lists this belief by name among the ideas that students bring to Unit 11: that current is used up in a circuit. It feels right because something in the bulb clearly is being used, the bulb is giving out light and heat, and current is the name for what flows into it.

Put an ammeter in the loop and the belief fails its first test. It reads the same current between the battery and the first bulb, between the two bulbs, and between the second bulb and the battery. And the two bulbs glow equally. This lesson traces exactly where the "used up" reasoning goes wrong and replaces it with a model that works. Then it uses that model on the brightness puzzles the exam likes to set, including two where the answer reverses.

Where the reasoning breaks: what the bulb takes

The broken step is the assumption that what the bulb uses is current. Current is a flow of charge, and charge is conserved. If charge flowed into a bulb faster than it flowed out, charge would pile up inside the bulb and its potential would rise. That would quickly push back on the incoming charge until the flows matched. In a steady circuit nothing piles up anywhere, so the current into any element equals the current out of it. In a single loop, the current is the same at every point.

What the bulb takes is energy. Each coulomb of charge leaving the battery has been raised through the battery's potential difference. As it passes through the bulb it falls through a potential difference ΔV, losing ΔV joules of electric potential energy, which the filament turns into light and thermal energy. The charge comes out of the bulb with less energy, not less of itself. A bicycle chain is a fair picture: the chain moves at the same rate at every link, while the energy goes from your legs to the rear wheel. Where the picture fails is that the charges in a circuit start out everywhere at once and there is no single link that carries the energy across.

The point: in a series loop the current is the same everywhere, and the potential difference is shared among the elements. Bulbs convert energy; they do not consume charge.

Series and parallel, side by side

A series connection is one in which any charge passing through one element must go through all of them, with no other path: one loop, one current. A parallel connection is one in which charge can flow through any one of two or more paths, each connected across the same two points. So each path has the same potential difference across it. Conservation of charge and of energy give the rules for each.

The equivalent resistance is the single resistance that would draw the same current from the same source. In series, each resistor adds to the obstruction, so R_eq is the sum. In parallel, every extra path is another way through, so the equivalent resistance falls and is always less than the smallest resistance in the group: 6.0 Ω and 3.0 Ω in parallel give 1/R_eq = 1/6.0 + 1/3.0 = 3/6.0, so R_eq = 2.0 Ω.

Power, and what brightness really measures

The rate at which an element converts electrical energy is its power. Each coulomb gives up ΔV joules and I coulombs pass each second, so

P = IΔV = I2R = ΔV2/R

where the last two forms follow from Ohm's law. Power is in watts, joules per second. A bulb's brightness increases with its power, so power is the quantity to rank when asked which bulb is brightest. Choose the form that matches what the elements share: in series they share the current, so P = I²R is convenient; in parallel they share the potential difference, so P = ΔV²/R is.

Worked problem: two bulbs, two arrangements

Givens. Two identical ohmic bulbs of 5.0 Ω each, and an ideal 3.0 V battery.

The set-up in words. First the bulbs are wired one after the other in a single loop with the battery. Then they are rewired so that each bulb is connected directly across the battery's two terminals.

Principle. Equivalent resistance, Ohm's law, then P = IΔV for each bulb.

1. Series: R_eq = 5.0 + 5.0 = 10 Ω, so I = (3.0 V)/(10 Ω) = 0.30 A through both bulbs. Each has ΔV = IR = 1.5 V and P = (0.30 A)(1.5 V) = 0.45 W.
2. Parallel: each bulb has the full 3.0 V, so each carries I = (3.0 V)/(5.0 Ω) = 0.60 A and dissipates P = (3.0 V)2/(5.0 Ω) = 1.8 W. The battery supplies 1.2 A in all.

Sanity check. In series the bulbs are equally bright, as the ammeter readings demanded, and the total power, 0.90 W, equals the battery's output, (3.0 V)(0.30 A). In parallel each bulb gets four times the power, because doubling ΔV doubles the current too, and the total, 3.6 W, again equals (3.0 V)(1.2 A).

The variant that flips the result. Now use two different bulbs, 5.0 Ω and 10 Ω. In series they share a current of (3.0 V)/(15 Ω) = 0.20 A, and P = I²R gives 0.20 W for the 5.0 Ω bulb and 0.40 W for the 10 Ω bulb: the higher resistance glows brighter. In parallel each has 3.0 V, and P = ΔV²/R gives 1.8 W for the 5.0 Ω bulb and 0.90 W for the 10 Ω bulb: now the lower resistance glows brighter. The same two bulbs swap places in the ranking when the wiring changes, which is why "higher resistance means brighter" and "lower resistance means brighter" are both wrong as general rules.

Worked problem: the bulb that dims when another goes out

Givens. Three identical ohmic bulbs of 6.0 Ω and an ideal 9.0 V battery. Bulb A is in series with a parallel pair, B and C.

The set-up in words. From the battery's positive terminal, the wire runs through bulb A, then divides into two branches, one through B and one through C, which rejoin and return to the battery.

Principle. Reduce the circuit step by step to one equivalent resistance, find the total current, then work back out to each bulb.

1. B and C in parallel: 1/R = 1/6.0 + 1/6.0, so R_BC = 3.0 Ω. With A in series: R_eq = 6.0 + 3.0 = 9.0 Ω.
2. Total current: I = (9.0 V)/(9.0 Ω) = 1.0 A, all of it through A.
3. Bulb A: ΔV_A = (1.0 A)(6.0 Ω) = 6.0 V and P_A = 6.0 W. The pair: ΔV_BC = (1.0 A)(3.0 Ω) = 3.0 V, so B and C each carry 0.50 A and dissipate (0.50 A)(3.0 V) = 1.5 W.

Sanity check. The potential differences around the loop add to the battery's: 6.0 V + 3.0 V = 9.0 V. The currents into the junction equal the currents out: 1.0 A = 0.50 A + 0.50 A. And the powers add to the battery's output: 6.0 + 1.5 + 1.5 = 9.0 W = (9.0 V)(1.0 A). A is brightest, and B and C are equal, each with a quarter of A's power.

The variant that flips the result. Unscrew bulb C. Now A and B are in series: R_eq = 12 Ω, the current is (9.0 V)/(12 Ω) = 0.75 A, and each bulb dissipates (0.75 A)²(6.0 Ω) = 3.4 W. Removing a bulb made A dimmer, from 6.0 W to 3.4 W, because the total resistance rose; and it made B brighter, from 1.5 W to 3.4 W, because B no longer shares the current. Anyone reasoning that "less load means everything gets brighter", or that "B is unaffected because C was on a separate branch", gets one of the two bulbs wrong.

The point: after any change, recompute from the outside in: new equivalent resistance, new total current, then each element. Local guesses about one branch fail because every part of a circuit affects the current through the rest.

A second belief worth retiring: the battery as a current supply

Paula Engelhardt and Robert Beichner published a study in 2004 of students taking a diagnostic test on DC circuits. It found that the idea students relied on most often in interviews was that a battery is a constant source of current. An ideal battery is nothing of the kind. It maintains a fixed potential difference, and the current it delivers is whatever the circuit draws. In the worked problem above, that was 0.30 A for two bulbs in series and 1.2 A for the same bulbs in parallel. That is why plugging more devices into one source in parallel increases the total current, and why a battery with many parallel loads runs flat sooner.

Common misconceptions

• "Current is used up by the first bulb." The current is the same everywhere in a series loop. Bulbs convert the charges' electric potential energy; the charge itself goes all the way round.
• "A battery always delivers the same current." It maintains a potential difference; the current depends on the circuit's resistance.
• "Adding resistors in parallel adds to the resistance." It lowers it, below the smallest resistance in the group, because each new path is another way through.
• "A higher-resistance bulb is always brighter." In series, yes, by P = I²R; in parallel the lower-resistance bulb is brighter, by P = ΔV²/R.
• "In parallel, the current always splits equally." Only for equal resistances; more current takes the path of lower resistance, since each path has the same ΔV.

Recap

Two identical bulbs in series glow equally, and an ammeter reads the same current before, between and after them. Charge is conserved and bulbs convert energy, not charge, so the "used up" model fails at its first step. In series the current is the same everywhere and the potential difference is divided, with R_eq equal to the sum of the resistances. In parallel the potential difference is the same on every path and the currents add, with 1/R_eq equal to the sum of the reciprocals, always below the smallest resistance.

Brightness follows power, P = IΔV = I²R = ΔV²/R. Two 5.0 Ω bulbs on 3.0 V give 0.45 W each in series and 1.8 W each in parallel, and a 10 Ω bulb beats a 5.0 Ω one in series but loses to it in parallel. With bulb A in series with a parallel pair on 9.0 V, A takes 6.0 W and B and C 1.5 W each. Unscrewing C dims A to 3.4 W and brightens B to 3.4 W. A battery fixes a potential difference, not a current, so the current it supplies rises as parallel loads are added.

Sources

1. OpenStax. (2022). 20.4 Electric power and energy. In College Physics 2e. openstax.org
2. OpenStax. (2022). 21.1 Resistors in series and parallel. In College Physics 2e. openstax.org
3. Engelhardt, P. V., and Beichner, R. J. (2004). Students' understanding of direct current resistive electrical circuits. American Journal of Physics, 72(1), 98-115. https://doi.org/10.1119/1.1614813. Preprint: arxiv.org
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 11.4 and 11.5, and the Unit 11 guide's note on the beliefs that batteries store charge and that current is used up. apcentral.collegeboard.org
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Circuit construction kit: DC (optional: build the circuits in this lesson and read every current). phet.colorado.edu

A cell that reads 1.58 V, then 1.38 V

Here is a measurement you can repeat with a multimeter. A fresh AA cell with nothing connected to it reads 1.58 V across its terminals. Clip a 2.0 Ω resistor across the same terminals, leaving the meter where it is, and the reading drops to 1.38 V, while a meter in series shows 0.69 A flowing. Nothing about the cell's chemistry changed when the resistor was attached, so where did 0.20 V go? And why would the reading drop even further, to about 0.8 V, if the resistor were replaced by a small motor of 0.30 Ω?

The answer needs one new idea about batteries and two rules that turn the conservation laws into circuit equations. Together they let you analyse any circuit the exam can set, and they explain why the instruments you use to measure a circuit must be designed so as not to change it.

The missing 0.20 V: internal resistance

The potential difference a battery maintains when no current flows through it is its emf, ε: 1.58 V for this cell. When current flows, the charges also have to pass through the battery's own materials, which oppose their motion. A real battery is therefore modelled as an ideal source of emf ε in series with a small internal resistance r. The potential difference you can actually measure across its terminals, the terminal voltage, is the emf minus the drop across that internal resistance:

ΔVterminal = ε - Ir

With no current, the terminal voltage equals the emf, which is why the meter reads 1.58 V with nothing connected. The more current the battery delivers, the more it loses inside itself. OpenStax adds that internal resistance rises as a battery is depleted, which is why lights left on in a parked car slowly dim as the battery runs down. An ideal battery has no internal resistance, and the exam assumes ideal batteries unless it says otherwise.

Givens. A cell with ε = 1.58 V and r = 0.29 Ω is connected to a load resistor R = 2.0 Ω.

The set-up in words. One loop: the ideal emf, the internal resistance just after it inside the cell's casing, and the external resistor, connected by wires of negligible resistance.

Principle. The internal resistance is in series with the load, so the current is the emf divided by the total resistance; then ΔV = ε - Ir.

1. Current: I = ε/(R + r) = (1.58 V)/(2.0 Ω + 0.29 Ω) = 0.69 A.
2. Terminal voltage: ΔV = ε - Ir = 1.58 V - (0.69 A)(0.29 Ω) = 1.58 V - 0.20 V = 1.38 V.
3. Power: the load receives I²R = 0.95 W and the cell's own interior I²r = 0.14 W. The load's share is R/(R + r) = 2.0/2.29 = 87 percent.

Sanity check. The 0.20 V lost inside the cell plus the 1.38 V across the resistor adds back to the 1.58 V emf. Running the measurement backwards gives the internal resistance from the opening readings: r = (1.58 V - 1.38 V)/(0.69 A) = 0.29 Ω.

The variant that flips the result. Replace the 2.0 Ω resistor with a 0.30 Ω motor. Now I = (1.58 V)/(0.59 Ω) = 2.7 A, and the terminal voltage collapses to 1.58 V - (2.7 A)(0.29 Ω) = 0.80 V. The load's share of the energy falls from 87 percent to 0.30/0.59 = 51 percent: almost as much power, 2.1 W, now heats the inside of the cell as drives the motor. For a gentle load the internal resistance is a small correction; for a heavy load it decides what happens, and it is why a battery connected straight across a thick wire, as in the electromagnet of Lesson 17, quickly gets hot.

The junction rule: charge in equals charge out

Where wires meet at a junction, charge cannot pile up, so Kirchhoff's junction rule says the total current flowing into a junction equals the total current flowing out:

ΣIin = ΣIout

It is conservation of charge, applied one junction at a time. If 2.0 A arrives along one wire and 0.5 A leaves along a second, the third wire must carry 1.5 A out. You met it in the last lesson without the name, when 1.0 A through bulb A split into 0.50 A through each of B and C.

The loop rule: the potential comes back to where it started

Potential is a property of position. If you trace any closed loop in a circuit and return to your starting point, you must arrive back at the potential you started from, so Kirchhoff's loop rule says the potential differences around any closed loop add to zero:

ΣΔV = 0

This is conservation of energy: a charge that goes round a loop gains energy qε crossing each battery from its negative to its positive terminal and loses energy qIR crossing each resistor in the direction of the current, and it must end with the energy it began with. The sign rules follow directly. Crossing a battery from - to +, the change is +ε; from + to -, it is -ε. Crossing a resistor in the direction of the current, the change is -IR; against the current, +IR.

Givens. A 9.0 V battery and a 6.0 V battery, both ideal, are connected in a single loop with a 1.0 Ω and a 2.0 Ω resistor. The batteries face each other: their positive terminals are connected through the 1.0 Ω resistor.

The set-up in words. One loop running from the negative terminal of the 9.0 V battery, through it, through the 1.0 Ω resistor, into the positive terminal of the 6.0 V battery, out of its negative terminal, and back through the 2.0 Ω resistor.

Principle. The loop rule, with a guessed direction of current; a negative answer would simply mean the guess was backwards.

1. Guess that the current flows the way the larger battery pushes it, out of the 9.0 V battery's positive terminal. Going round in that direction: +9.0 V - I(1.0 Ω) - 6.0 V - I(2.0 Ω) = 0.
2. Solve: 3.0 V = I(3.0 Ω), so I = 1.0 A, positive, so the guess was right.

Sanity check. Read it as a graph of potential against position around the loop, starting at 0 V at the negative terminal of the 9.0 V battery. The potential steps up 9.0 V across that battery, slopes down 1.0 V across the 1.0 Ω resistor to 8.0 V, steps down 6.0 V crossing the 6.0 V battery from + to - to 2.0 V, and slopes down the last 2.0 V across the 2.0 Ω resistor, arriving back at 0 V where it began. The 6.0 V battery is being pushed backwards through itself: it is being charged by the larger one.

The variant that flips the result. Turn the 6.0 V battery round so that the two batteries aid each other. Now +9.0 V + 6.0 V - I(3.0 Ω) = 0 and I = 5.0 A, five times as much, and the smaller battery is now discharging, supplying energy instead of absorbing it. One reversed component changes both the size of the current and the job each battery does.

Two loops, worked with both rules

A 12 V ideal battery drives a 4.0 Ω resistor, after which the current divides between a 6.0 Ω resistor and a 3.0 Ω resistor in parallel. Call the currents I₁ through the 4.0 Ω resistor, I₂ through the 6.0 Ω and I₃ through the 3.0 Ω.

1. Junction rule where the current divides: I1 = I2 + I3.
2. Loop through the battery, the 4.0 Ω and the 6.0 Ω: 12 - 4.0I1 - 6.0I2 = 0.
3. Loop round the two parallel resistors alone: 6.0I2 - 3.0I3 = 0, so I₃ = 2I₂, and then I₁ = 3I₂.
4. Substitute into the first loop: 12 - 12I₂ - 6.0I₂ = 0, so I2 = 0.67 A, I3 = 1.33 A and I1 = 2.0 A.

Check with the last lesson's method: the parallel pair is 2.0 Ω, so R_eq = 6.0 Ω and I₁ = 12 V/6.0 Ω = 2.0 A. The rules give the same answer, and they keep working for circuits that cannot be reduced to series and parallel pieces.

In short: Kirchhoff's rules are the conservation of charge (junctions) and of energy (loops), written as equations. Choose current directions, write one junction equation and enough loop equations, and solve.

Meters that do not disturb

An ammeter measures the current passing through it, so it must be connected in series, in the same branch as the element whose current you want. Connecting an ammeter across a component puts a very low resistance in parallel with it, which short-circuits it and can blow the meter's fuse. An ideal ammeter has zero resistance, so it does not change the current it measures. A voltmeter measures the potential difference between the two points it touches, so it is connected in parallel, across the component. An ideal voltmeter has infinite resistance, so no current is diverted through it.

Real meters fall short of both ideals, and the exam asks you to say qualitatively what that does. An ammeter with some resistance adds to the resistance of the loop it sits in, so it reads slightly less current than flowed before it was inserted: a 0.50 Ω ammeter in a loop with a 2.0 Ω resistor and an ideal 1.5 V battery reads 0.60 A instead of 0.75 A. A voltmeter of finite resistance provides a parallel path, lowering the resistance of the part it is connected across and therefore its share of the voltage: two 10 kΩ resistors in series on 9.0 V each have 4.5 V, but a 10 kΩ voltmeter across one of them turns that part into 5.0 kΩ and reads only 3.0 V. As OpenStax notes, a digital voltmeter's resistance can be far greater than an older analog meter's, so it disturbs ordinary circuits very little.

Common misconceptions

• "A battery always delivers its rated voltage." Its terminal voltage is ε - Ir and falls as the current rises: 1.38 V at 0.69 A from a 1.58 V cell, 0.80 V at 2.7 A.
• "Emf is a force." Despite its old name, it is a potential difference, measured in volts.
• "An ammeter is connected across the component." That is how a voltmeter is connected. An ammeter goes in series; across a component it short-circuits it.
• "Kirchhoff's rules are new laws to memorise." They are the conservation of charge and of energy, applied to junctions and loops.
• "Wires can always be ignored." Only when the circuit has other, larger resistances. A wire straight across a battery is limited only by the battery's internal resistance, and it gets hot.

What to carry forward

A fresh AA cell that reads 1.58 V on open circuit gives 1.38 V to a 2.0 Ω resistor because 0.20 V is lost across its internal resistance of 0.29 Ω, and only 0.80 V to a 0.30 Ω motor, which then gets just half the cell's energy. A real battery is an ideal emf in series with an internal resistance, with terminal voltage ΔV = ε - Ir; the emf is the reading with no current, and the internal resistance grows as the battery is used up. Kirchhoff's junction rule, current in equals current out, is conservation of charge; his loop rule, potential differences around a closed loop sum to zero, is conservation of energy, and a graph of potential against position around a loop rises at each battery crossed from - to + and falls across each resistor, returning to its start. Two opposing batteries of 9.0 V and 6.0 V through 3.0 Ω drive 1.0 A and charge the smaller one; turned to aid each other they drive 5.0 A. Ammeters go in series and ideally have zero resistance; voltmeters go in parallel and ideally have infinite resistance; real ones read slightly low, an effect the exam asks you to describe in words.

Sources

1. OpenStax. (2022). 21.2 Electromotive force: Terminal voltage. In College Physics 2e. openstax.org
2. OpenStax. (2022). 21.3 Kirchhoff's rules. In College Physics 2e. openstax.org
3. OpenStax. (2022). 21.4 DC voltmeters and ammeters. In College Physics 2e. openstax.org
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 11.5, 11.6 and 11.7 (non-ideal meters treated qualitatively; no parallel batteries of different emf). apcentral.collegeboard.org

A voltmeter that climbs and never quite arrives

Connect an uncharged 1000 μF capacitor in series with a 10 kΩ resistor and a 3.0 V battery, with a voltmeter across the capacitor. At the instant the circuit closes the meter reads 0 V. Ten seconds later it reads 1.9 V. After twenty seconds, 2.6 V. After fifty, 2.98 V. It never quite reaches 3.0 V, although after a minute nobody could tell the difference. Disconnect the battery, put the resistor straight across the charged capacitor, and the reading falls the same way in reverse: fast at first, then ever more slowly.

The exam does not ask you to write the curve's equation. It asks you to describe the circuit at three moments, the first instant, one time constant later, and the long run, and to compare charging with discharging and capacitors with resistors. Those comparisons are the backbone of this lesson, so it is built around two tables.

Comparison one: capacitors combine the opposite way to resistors

Capacitors in a circuit can be replaced by a single equivalent capacitance, just as resistors can, but the rules are swapped. Capacitors in parallel are effectively one capacitor with a larger plate area, so their capacitances add. Capacitors in series all carry the same charge, because the plates joined between them start neutral and can only separate charge, never gain it. Their potential differences add, and the equivalent capacitance is less than the smallest of them.

Take 3.0 μF and 6.0 μF. In series, 1/C_eq = 1/3.0 + 1/6.0 = 3/6.0, so C_eq = 2.0 μF. Across 12 V, each carries Q = (2.0 μF)(12 V) = 24 μC. So the 3.0 μF capacitor has Q/C = 8.0 V and the 6.0 μF has 4.0 V, adding to 12 V. Notice the smaller capacitor takes the larger share of the voltage. In parallel, C_eq = 9.0 μF; each has the full 12 V, and they hold 36 μC and 72 μC.

The time constant

When a capacitor charges or discharges through a resistance, how quickly it does so is set by one number, the time constant:

τ = ReqCeq

The units work out to seconds: an ohm is a volt per ampere and a farad a coulomb per volt, so Ω·F = C/A = s. For a charging capacitor, τ is the time for its charge, and its voltage, to rise from zero to about 63 percent of the final value. For a discharging capacitor, τ is the time for its charge to fall to about 37 percent of the starting value. The physical reason for the slowing is simple. A charging capacitor's own voltage opposes the battery, so as it fills, the voltage left across the resistor shrinks and the current with it; less current means slower filling. A discharging capacitor's voltage drives its own current, so as it empties, its push weakens.

Comparison two: the first instant, one time constant, the long run

For a capacitor C charging through a resistor R from an ideal battery of emf ε, and for the same capacitor, once full, discharging through R:

Read the first and last columns first, because they carry most exam questions. At the first instant an uncharged capacitor has no charge and so no voltage across it, ΔV = Q/C = 0. It behaves exactly like a wire, and the whole battery voltage sits across the resistor, driving the largest current the circuit will ever have. After a long time the capacitor is full and the circuit has reached a steady state. The capacitor's voltage matches the battery's, nothing is left to push charge through the resistor, and the current in that branch is zero. The capacitor behaves like a break in the circuit.

Then read the middle column. Energy grows as the square of the charge, so at 63 percent of the charge the capacitor holds only 0.63² = 40 percent of its final energy. In discharge, 37 percent of the charge leaves only 0.37² = 14 percent of the energy. And notice that every quantity approaches its final value without arriving: after 5τ the charge is within 1 percent of full.

The core of it: an uncharged capacitor is a wire at the first instant, and a full one is an open switch in the long run. The time constant τ = RC says how long the change takes.

Worked problem: a series RC circuit, charged and then discharged

Givens. An ideal 3.0 V battery, a 10 kΩ resistor and an uncharged 1000 μF capacitor in series, with a switch.

The set-up in words. A single loop: battery, switch, resistor, capacitor, back to the battery. A voltmeter is connected across the capacitor.

Principle. The loop rule at each moment, ε = IR + Q/C, together with the first-instant and long-run behaviour, and τ = RC.

1. Time constant: τ = RC = (1.0 × 104 Ω)(1.0 × 10-3 F) = 10 s.
2. First instant: Q = 0, so the loop rule gives ε = IR, and I0 = (3.0 V)/(1.0 × 104 Ω) = 3.0 × 10-4 A, 0.30 mA.
3. Long run: I = 0, so Q/C = ε: Q = Cε = (1.0 × 10-3 F)(3.0 V) = 3.0 × 10-3 C, storing U = (1/2)Cε2 = 4.5 × 10-3 J.
4. After one time constant, 10 s: ΔV_C is about 0.63 × 3.0 V = 1.9 V and the current about 0.37 × 0.30 mA = 0.11 mA, as the opening readings showed.

Sanity check. At every moment the loop rule holds: at 10 s, IR = (0.11 mA)(10 kΩ) = 1.1 V and Q/C = 1.9 V, which add to the battery's 3.0 V. And the energy accounts reveal something surprising. The battery moves 3.0 × 10^-3 C through 3.0 V, doing 9.0 × 10^-3 J of work, but the capacitor ends up storing only half of that. The other half was converted to thermal energy in the resistor, and that is true whatever the value of R.

The variant that flips the result. Now remove the battery and connect the resistor straight across the full capacitor. The capacitor becomes the source. At the first instant it drives 3.0 V/10 kΩ = 0.30 mA, but in the opposite direction round the loop to the charging current. Charge now flows off the plates instead of onto them. After 10 s the voltage has fallen to about 1.1 V, 37 percent, and the stored energy to 14 percent of 4.5 mJ. Change the resistance instead, say to 20 kΩ, and the time constant doubles to 20 s and the first-instant current halves. But the final charge is still 3.0 × 10^-3 C: the resistance decides how fast, never how much.

A branch that waits: resistors and a capacitor together

The exam's favourite RC circuit puts the capacitor in parallel with a resistor. An ideal 12 V battery drives current through R₁ = 2.0 kΩ, after which the circuit divides between R₂ = 4.0 kΩ and an uncharged 50 μF capacitor, which are in parallel.

At the first instant, the capacitor acts like a wire, so the voltage across it is zero, and therefore so is the voltage across R₂, which is in parallel with it. No current flows through R₂. All of it flows through R₁ and into the capacitor: I = 12 V/2.0 kΩ = 6.0 mA.

After a long time, the capacitor is an open switch and carries no current, so the battery drives current through R₁ and R₂ in series: I = 12 V/6.0 kΩ = 2.0 mA. The capacitor's voltage equals R₂'s, (2.0 mA)(4.0 kΩ) = 8.0 V, so it holds Q = (50 μF)(8.0 V) = 400 μC and stores (1/2)(50 μF)(8.0 V)² = 1.6 mJ. In between, R₂'s current grows from 0 to 2.0 mA while the capacitor's branch current falls from 6.0 mA to 0.

Open the battery's switch after that, and the capacitor discharges through R₂ alone, since R₁ is in the broken branch. The current through R₂ starts at 8.0 V/4.0 kΩ = 2.0 mA and dies away with time constant R₂C = (4.0 kΩ)(50 μF) = 0.20 s. Meanwhile the current in the capacitor's own branch has reversed.

Common misconceptions

• "Current flows through a capacitor." No charge crosses the gap. Charge flows onto one plate and off the other, so current flows in the wires on both sides while the plates fill, and stops when they are full.
• "An uncharged capacitor blocks current at first." It does the opposite: with no voltage across it, it behaves like a wire, and the current is largest at the first instant.
• "The time constant is the time to charge fully." It is the time to reach about 63 percent. A capacitor approaches full charge without ever arriving exactly; five time constants gets within 1 percent.
• "A larger resistance means less charge is stored." The final charge in a series RC circuit is Cε, independent of R. Resistance only slows the process.
• "Capacitors in series add up like resistors in series." Their reciprocals add, so the equivalent capacitance is less than the smallest one.

What to remember

A 1000 μF capacitor charged through 10 kΩ from 3.0 V reaches 1.9 V after 10 s and creeps toward 3.0 V without arriving. Capacitors combine opposite to resistors. In parallel they add, 3.0 μF and 6.0 μF making 9.0 μF. In series their reciprocals add, making 2.0 μF, each carrying the same 24 μC on 12 V, with the smaller capacitor taking the larger voltage. The time constant τ = R_eqC_eq, 10 s here, is the time to reach about 63 percent of the final charge when charging. When discharging, it is the time to fall to about 37 percent.

At the first instant an uncharged capacitor is a wire and the current is largest, ε/R. In the long run a full one is an open switch, with no current in its branch and the battery's voltage across it, Q = Cε. Charging through a resistor always costs as much energy in the resistor as ends up stored. With a capacitor in parallel with R₂, R₂ carries nothing at first and its full share in the long run. When the battery is removed, the capacitor discharges through R₂ with τ = R₂C.

Sources

1. OpenStax. (2022). 21.6 DC circuits containing resistors and capacitors (the time constant; 63 and 37 percent). In College Physics 2e. openstax.org
2. OpenStax. (2022). 19.6 Capacitors in series and parallel. In College Physics 2e. openstax.org
3. OpenStax. (2022). 19.7 Energy stored in capacitors. In College Physics 2e. openstax.org
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 11.8, Resistor-capacitor (RC) circuits (qualitative descriptions; initial and final states treated mathematically). apcentral.collegeboard.org
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Circuit construction kit: DC (optional: the full version includes capacitors). phet.colorado.edu

A fridge magnet snapped in two

Snap a ceramic fridge magnet in half and you might expect to be holding a north pole in one hand and a south pole in the other. You are not. Each half has a north pole and a south pole of its own, and each will cling to the fridge door. Snap one half again and the same thing happens. Keep going, in imagination, past the point where a knife could help: through the microscopic regions of a magnetized material, down to single atoms, down to the electrons inside them. At every stage the pieces are still two-poled magnets, called magnetic dipoles. Nobody has ever found a piece with only one pole, an isolated north or south pole called a magnetic monopole, although physicists keep looking.

This lesson follows that chain of dipoles downward, to find where magnetism comes from, and then back up to the largest magnet you will ever stand on, the Earth, whose own dipole is the reason a compass works.

The magnetic field, drawn with a compass

A magnet changes the space around it: a compass needle placed nearby swings round and settles in a particular direction. That direction, at each point, is the direction of the magnetic field, B, a vector field that determines the magnetic force on moving charges, on electric currents and on magnetic materials. By definition, the field points the way the north end of a compass needle points. Its SI unit is the tesla (T). For a sense of scale, NOAA's geomagnetism service gives the strength of the Earth's field at the surface as roughly 25 000 to 65 000 nanotesla, that is 25 to 65 μT. Close to even a small fridge magnet the field is many times stronger, as the activity at the end of this lesson lets you measure.

Move a compass around a bar magnet and trace the direction it points, and you can draw magnetic field lines, curves that follow the field everywhere, as with electric fields in Lesson 8. Outside the magnet they leave the north pole, sweep round through the space beside it, and enter the south pole. They do not stop there. Inside the magnet they continue from the south pole back to the north, so every line is a closed loop with no beginning and no end. That is the deep difference from electric field lines, which start on positive charges and end on negative ones: there are no magnetic charges for the lines to start or end on. A field-vector map of a bar magnet shows the same picture as arrows, longest near the poles, where the lines crowd together, and shortest far away.

A compass needle is itself a small dipole, and a dipole placed in a magnetic field turns to line up with it. That is all a compass does: it reports the direction of the net field where it sits.

Poles come in pairs, and the field weakens fast

The behaviour of poles echoes that of charges in some ways and not in others. Like poles repel and unlike poles attract, just as like charges repel. But an electric charge can sit on its own, while a magnetic pole never does: north and south always come together, which is why the snapped fridge magnet gives two complete magnets.

The field of a dipole also weakens with distance faster than the field of a single charge. Close to the magnet the details of its shape matter, but far away, many times its own length, the field falls off as 1/r³, as OpenStax's University Physics shows for the field on the axis of a small current loop, the simplest dipole. Double the distance and the field drops to an eighth. That is why a magnet that does nothing to a paper clip 5 cm away grabs it suddenly at 1 cm.

Where the dipoles come from: moving charge

OpenStax states the source of all magnetism in a single sentence: electric current is the source of all magnetism. A loop of current makes a dipole field, and so does anything in which charge circulates. Inside an atom, electrons orbit the nucleus and also have an intrinsic spinning property, and each of these motions makes the electron a tiny magnetic dipole. In most atoms these dipoles pair off and cancel, and in most materials the atoms' leftover dipoles point every which way and cancel each other too.

In a few elements, iron, cobalt, nickel and gadolinium, the atoms' dipoles do not cancel, and neighbouring atoms strongly prefer to line their dipoles up with each other. Whole regions of the material, called magnetic domains, end up with their dipoles aligned. In an unmagnetized piece of iron, the domains are small and point in random directions, so their fields cancel. Put the iron in an external field and the domains aligned with the field grow at the expense of the others, reaching millimetre size according to OpenStax, and the iron becomes a magnet. Permanent magnetism and induced magnetism are both this alignment of dipoles; the difference is whether the alignment survives when the external field is removed.

Heat breaks the alignment. Above a temperature called the Curie temperature, thermal agitation scrambles the domains and the material cannot be magnetized: for iron, OpenStax gives 1043 K, or 770 °C. A magnet heated above that and cooled with no field present comes back unmagnetized.

Three ways a material can respond

A material's magnetic permeability measures how strongly it becomes magnetized in response to an external field. Empty space has the constant value μ₀, the vacuum permeability, which appears in every equation for the field of a current. The exam's table gives μ₀ = 4π × 10^-7 T·m/A. Before the 2019 revision of the SI that value was exact by definition; now μ₀ is measured, and the CODATA 2022 value is 1.256 637 061 27 × 10^-6 N/A², which agrees with 4π × 10^-7 to about one part in ten billion. Matter is different: HyperPhysics notes that the relative permeability of paramagnetic and diamagnetic materials is very close to 1, while for ferromagnetic materials it can be very large. And unlike μ₀, the permeability of a material is not a fixed number: it depends on temperature, on orientation and on the strength of the applied field.

The induced magnetism of a ferromagnet explains why a magnet picks up a paper clip that was never magnetized. The magnet's field aligns the clip's domains so that its nearer end becomes an opposite pole, and attraction follows, much as a charged strip polarizes a scrap of paper.

Worked problem: a compass caught between two fields

Givens. At a compass on a table, the horizontal component of the Earth's field is 20 μT, pointing north. A bar magnet is laid on the table due east of the compass, along the east-west line, and at the compass its field is 20 μT, pointing east.

The set-up in words. Looking down on the table: the compass in the middle, north at the top of the page, and the bar magnet to the right, lying horizontally.

Principle. Magnetic fields add as vectors, and the needle aligns with the net field.

1. Components of the net field: 20 μT north and 20 μT east.
2. Magnitude: B = √(202 + 202) μT = 28 μT.
3. Direction: θ = tan-1(20/20) = 45° east of north. The needle points north-east.

Sanity check. Equal perpendicular components must give a direction halfway between them, and the magnitude must exceed either component but be less than their sum.

The variant that flips the result. Turn the magnet end for end. Its field at the compass now points west, and the needle swings to 45° west of north, the mirror image. Alternatively, slide the magnet out to twice its original distance. Far from the magnet its field falls as 1/r³, so it drops to 20/8 = 2.5 μT, and the needle turns only tan^-1(2.5/20) = 7° east of north. A compass is a field meter: the angle it turns tells you the ratio of the magnet's field to the Earth's.

What matters here: the direction of a magnetic field at a point is the direction a compass points there, and fields from several sources add as vectors.

The Earth is a dipole, upside down

NOAA describes the Earth's field as very similar to the field a dipole magnet at the Earth's centre would produce, and attributes more than 90 percent of it to the Earth's outer core, where moving molten iron acts as a dynamo: circulating charge, once again. The field lines emerge from the southern hemisphere, arch over the equator, and plunge into the ground in the far north. NOAA notes that at the north magnetic pole the north end of a dip needle, a compass free to tilt vertically, points straight down.

That settles an old puzzle. The north end of a compass is attracted to the Arctic, and unlike poles attract, so the magnetic pole in the Arctic must be a south pole of the Earth's dipole. It is called the north magnetic pole because of where it is, not what it is. The exam treats the Earth's field as that of a dipole, and expects you to know that a compass aligns with it.

Common misconceptions

• "Breaking a magnet separates its north and south poles." Each piece is a complete dipole. No isolated pole has ever been observed.
• "The Earth's north magnetic pole is a magnetic north pole." It attracts the north end of a compass, so it is magnetically a south pole; the field points down into the ground there.
• "Magnets attract all metals." Only ferromagnetic materials are strongly attracted. Aluminium and copper coins are not, although aluminium is very weakly paramagnetic.
• "Magnetic field lines start at the north pole and end at the south pole." They continue through the magnet and form closed loops.
• "A magnet's pull is about the same at any distance." A dipole's field falls off as 1/r³ far away: twice as far, an eighth as strong.

The short version

A snapped fridge magnet gives two complete magnets, and so would any piece down to a single atom: magnetism comes in dipoles, never in isolated poles. The magnetic field is the vector field a compass needle lines up with, measured in teslas; the Earth's is 25 to 65 μT at the surface. Its field lines leave a north pole, enter a south pole and close through the magnet, and far from a dipole the field falls as 1/r³. All magnetism comes from moving charge: electrons in atoms are tiny dipoles, and in iron, nickel, cobalt and gadolinium they align in domains, which an external field can line up to make a permanent or an induced magnet, until heating past the Curie temperature, 770 °C for iron, scrambles them. Materials are ferromagnetic, paramagnetic (aluminium, titanium, magnesium) or diamagnetic (all of them, weakly), and their permeability, unlike the measured constant μ₀ = 1.256 637 061 27 × 10^-6 N/A², varies with conditions. Fields add as vectors: 20 μT north and 20 μT east send a compass to 45° east of north. The Earth is close to a dipole whose south pole sits in the Arctic, which is why compasses point north.

Sources

1. OpenStax. (2022). 22.2 Ferromagnets and electromagnets (domains; Curie temperature of iron; current as the source of all magnetism). In College Physics 2e. openstax.org
2. OpenStax. (2022). 22.3 Magnetic fields and magnetic field lines. In College Physics 2e. openstax.org
3. OpenStax. (2016). 12.4 Magnetic field of a current loop (the far field of a dipole falls off as distance cubed). In University Physics Volume 2. openstax.org
4. National Centers for Environmental Information, NOAA. (n.d.). Geomagnetism frequently asked questions. ncei.noaa.gov
5. Nave, R. (n.d.). Magnetic properties of solids. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
6. National Institute of Standards and Technology. (2024). CODATA 2022 value: vacuum magnetic permeability; and (n.d.) SI redefinition (in effect from 20 May 2019). physics.nist.gov
7. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 12.1, Magnetic fields. apcentral.collegeboard.org

Electrons that follow the field down to the poles

The aurora forms between 80 and 500 km above the Earth's surface, NOAA's Space Weather Prediction Center explains. It appears when electrons collide with oxygen and nitrogen in the upper atmosphere and make them give off light. Those electrons are accelerated far out in the night side of the Earth's magnetic environment. They reach the atmosphere in a particular way: they follow the Earth's magnetic field down to the polar regions. That is why auroras crown the poles instead of lighting the whole sky.

Why should a charged particle follow a magnetic field line, rather than cross it? The answer is in the peculiar direction of the magnetic force, which is always at right angles both to the field and to the particle's own motion. This lesson works out that force and its consequences step by step. It starts with the force on one proton and the circle it is bent into, then an electron that circles the other way. Last comes a device that uses electric and magnetic forces together to sort particles by speed.

The force, and the right-hand rule

A magnetic field exerts no force on a charge at rest. On a charge q moving with velocity v through a field B, it exerts a force of magnitude

FB = qvB sinθ

where θ is the angle between v and B. The force is largest when the motion is perpendicular to the field (sin 90° = 1) and zero when the motion is along the field or against it (sin 0° = sin 180° = 0). The exam asks for calculations only at those three angles, and for qualitative reasoning in between. The equation also defines the unit of field: one tesla is one newton per ampere-metre. OpenStax gives the Earth's surface field as about 5 × 10^-5 T, the strongest permanent magnets as roughly 2 T, and superconducting electromagnets as 10 T or more.

The direction of the force is perpendicular to both v and B, which leaves two possibilities, and the right-hand rule picks between them. OpenStax's version is right-hand rule 1. Point the thumb of your right hand along the velocity v and your straight fingers along the field B, and the force on a positive charge comes straight out of your palm. For a negative charge, the force is exactly opposite, out of the back of your hand. Because the force is always perpendicular to the velocity, it does no work: it can change a charge's direction but never its speed or kinetic energy.

Step one: a proton bent into a circle

Givens. A proton, mass 1.67 × 10^-27 kg and charge +1.60 × 10^-19 C, moves at 2.0 × 10⁶ m/s to the right, into a uniform 0.50 T field that points into the page.

The set-up in words. Picture the page as the plane of motion, with the field running straight into it everywhere, like rain falling onto the page from above. The proton enters from the left edge moving horizontally.

Principle. F = qvB at θ = 90°, with the direction from the right-hand rule. Because the force is perpendicular to the velocity and constant in size, the motion is uniform circular motion, with the magnetic force supplying the centripetal force mv²/r.

1. Size of the force: F = (1.60 × 10-19 C)(2.0 × 106 m/s)(0.50 T) = 1.6 × 10-13 N.
2. Direction at the moment of entry: thumb to the right (v), fingers into the page (B), palm facing the top of the page. The force points up the page, so the proton begins to curve upward, and the circle it follows turns anticlockwise.
3. Radius: qvB = mv2/r, so r = mv/(qB) = (1.67 × 10-27 kg)(2.0 × 106 m/s)/[(1.60 × 10-19 C)(0.50 T)] = 0.042 m, about 4 cm.
4. Time for one full circle: T = 2πr/v = 2π(0.042 m)/(2.0 × 106 m/s) = 1.3 × 10-7 s, the same as 2πm/(qB), which does not depend on the speed at all.

Sanity check. The units: kg·(m/s) divided by C·T, with T = N/(A·m) = N·s/(C·m), gives metres. The acceleration, F/m = 9.6 × 10¹³ m/s², is enormous, but so it should be for a particle this light under a force of this size. The speed stays at 2.0 × 10⁶ m/s all the way round, because the magnetic force does no work.

The variant that flips the result. Send an electron in with the same velocity. The force is the same size per unit charge but opposite in direction, out of the back of the hand, so the electron curves downward and circles clockwise. And since r = mv/(qB) and the electron's mass is 1836 times smaller, its circle has a radius of only 0.042 m/1836 = 23 micrometres. Change the direction of motion instead, so that the proton travels straight into the page, parallel to the field: then θ = 0°, the force is zero, and it goes straight on. Combine the two cases, a velocity partly along the field and partly across it, and the particle circles around the field line while sliding along it: a helix. That is how auroral electrons spiral down the Earth's field lines toward the poles.

The upshot: a magnetic force bends paths but never changes speeds. Perpendicular to the field, a charge circles with r = mv/(qB); along the field, it feels nothing.

Sorting by mass: the idea of a mass spectrometer

Because r = mv/(qB), ions with the same charge and speed in the same field follow circles whose radii are proportional to their masses. Singly charged carbon-12 and carbon-14 ions moving at 1.0 × 10⁵ m/s in a 0.20 T field travel on circles of radius 6.2 cm and 7.3 cm, so after half a circle they land about 2 cm apart. Measuring where ions land separates isotopes that no chemical method can tell apart. The difficulty is giving every ion the same speed, and that is the job of the next device.

Step two: crossed fields that pick one speed

Givens. A beam of positive ions passes between parallel plates that produce an electric field E = 1.0 × 10⁴ V/m pointing down the page. A magnetic field B = 0.010 T points into the page in the same region. The ions move to the right.

The set-up in words. The beam travels horizontally between two horizontal plates, the upper one positive, inside a magnet whose field runs into the page.

Principle. In a region with both fields, a charge feels both forces independently. The electric force qE does not depend on speed; the magnetic force qvB does.

1. Electric force on a positive ion: along E, down the page, size qE.
2. Magnetic force: thumb right, fingers into the page, palm up: up the page, size qvB.
3. No deflection when they balance: qE = qvB, so v = E/B = (1.0 × 104 V/m)/(0.010 T) = 1.0 × 106 m/s.

Sanity check. The charge cancels, and so does the mass, which never entered: ions of any charge and any mass pass straight through if, and only if, their speed is E/B. That is why this arrangement is called a velocity selector, and why a mass spectrometer often places one ahead of its deflecting magnet. The units: (V/m)/T = (N/C)/(N·s/(C·m)) = m/s.

The variant that flips the result. An ion moving faster than 1.0 × 10⁶ m/s feels the same electric force but a larger magnetic force, so it is deflected up the page; one moving slower is deflected down the page. Reverse the charge, a beam of electrons, and both forces reverse together, so the balance speed is unchanged and the deflections simply swap sides.

The Hall effect: a sideways voltage that reveals the carriers

The same sideways push acts on the charges drifting along a wire. Pass a current along a flat metal strip whose face is perpendicular to a magnetic field, and the magnetic force pushes the moving charge carriers toward one edge of the strip. Charge piles up on that edge, leaving the opposite edge with the opposite charge, until the electric field of the separated charges balances the magnetic force on the carriers. The result, described in OpenStax, is a small, steady potential difference across the width of the strip, the Hall emf, which is proportional to the field. Its sign depends on whether the carriers are positive or negative, so measuring it reveals which kind carries the current: it is how experiments showed that electrons carry current in metals while positive carriers do the job in some semiconductors. Small Hall probes built on the same effect are used to map magnetic fields.

Moving charges make magnetic fields too

The force law has a partner. A single moving charge is itself a source of magnetic field. At a point near its path, the field it produces is perpendicular both to its velocity and to the line from the charge to that point; it is strongest for points beside the path, where those two directions are perpendicular, and it weakens with distance. Its direction follows a curled right hand: point the thumb along the velocity of a positive charge, and the fingers curl round the path the way the field circles it. Many moving charges in a wire make the field of a current, the subject of the next lesson.

Common misconceptions

• "A magnetic field pushes a charge along the field lines." The force is perpendicular to the field and to the velocity. A charge moving along a field line feels no force at all.
• "The magnetic force speeds a particle up." It does no work, because it is always perpendicular to the motion. Speed and kinetic energy stay constant.
• "A stationary charge in a magnetic field feels a force." With v = 0, F = qvB sinθ = 0.
• "Electrons and protons in the same field curve the same way." Opposite charges curve in opposite directions, and the lighter electron on a far smaller circle.
• "A faster particle in a magnetic field circles more tightly." The radius r = mv/(qB) grows with speed, although the time for one circle does not change.

Where this leaves us

Auroral electrons follow the Earth's field lines down to the poles because of the direction of the magnetic force, F = qvB sinθ. It acts only on moving charges, and only at right angles both to the field and to the motion. It is zero along a field line, largest across it, and never able to change a particle's speed. Right-hand rule 1, thumb along v and fingers along B, puts the force on a positive charge out of the palm. The force on a negative charge comes out of the back of the hand.

A proton at 2.0 × 10⁶ m/s across a 0.50 T field feels 1.6 × 10^-13 N and circles with r = mv/(qB) = 4.2 cm. An electron at the same velocity circles the other way with a radius of 23 μm, and a velocity partly along the field gives a helix. Radii proportional to mass let a mass spectrometer separate isotopes. Crossed fields pass only particles with v = E/B, 1.0 × 10⁶ m/s for 1.0 × 10⁴ V/m and 0.010 T, deflecting faster particles one way and slower ones the other. The Hall effect, a sideways voltage across a current-carrying strip in a field, reveals the sign of the charge carriers. And moving charges are themselves sources of magnetic field, circling their path in the direction of a curled right hand.

Sources

1. NOAA Space Weather Prediction Center. (n.d.). Aurora. spaceweather.gov
2. OpenStax. (2022). 22.4 Magnetic field strength: Force on a moving charge in a magnetic field (right-hand rule 1; typical field strengths). In College Physics 2e. openstax.org
3. OpenStax. (2022). 22.5 Force on a moving charge in a magnetic field: Examples and applications. In College Physics 2e. openstax.org
4. OpenStax. (2022). 22.6 The Hall effect. In College Physics 2e. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 12.2, Magnetism and moving charges (quantitative force calculations at 0, 90 and 180 degrees only). apcentral.collegeboard.org

A compass beside a wire

In 1820, Hans Christian Oersted discovered that a current in a wire affected a compass needle placed near it. Electricity and magnetism had been studied for centuries as separate subjects, and here was a current, with no magnet anywhere, turning a magnet. Here is a version of his observation with numbers. A long straight wire runs north-south, 2.0 cm above a compass resting on a table. With no current, the needle points north along the wire. Switch on a steady current of 5.0 A flowing north, and the needle swings sharply. Which way does it swing, and how far? Answering that takes one formula, one right-hand rule and a vector sum, and the same three tools then explain the forces between wires and the workings of an electromagnet you can build with a nail.

The field of a long straight wire

A current is a stream of moving charges, and each moving charge makes a magnetic field, as the last lesson ended. Added up along a long straight wire, the result is a field whose lines are circles centred on the wire. The field at any point is tangent to the circle through that point: it has no component toward or away from the wire, and none along it. Its magnitude at a perpendicular distance r from the axis of the wire is

B = μ0I/(2πr)

proportional to the current and inversely proportional to the distance: twice as far away, half the field. That 1/r fall-off is gentler than the 1/r² of a point charge or the 1/r³ of a dipole. The direction comes from OpenStax's right-hand rule 2: point the thumb of your right hand in the direction of the current, and your fingers curl in the direction of the magnetic field loops.

Solving the opening problem

Givens. A long straight wire carries I = 5.0 A northward, 2.0 cm above a compass. The horizontal component of the Earth's field at the compass is 20 μT, pointing north. Use μ₀ = 4π × 10^-7 T·m/A.

The set-up in words. Seen from above, the wire lies along the north-south line directly over the compass. Seen end-on, looking north along the wire, the compass sits directly below it.

Principle. The wire's field from B = μ₀I/(2πr), its direction from right-hand rule 2, and the vector sum with the Earth's field, along which the needle settles.

1. Size of the wire's field at the compass: B = (4π × 10-7 T·m/A)(5.0 A)/[2π(0.020 m)] = 5.0 × 10-5 T = 50 μT.
2. Direction: thumb pointing north along the wire; the fingers curl round it. Looking north, the field circles clockwise, so directly below the wire it points west.
3. Net horizontal field: 20 μT north plus 50 μT west. The needle settles at θ = tan-1(50/20) = 68° west of north.

Sanity check. T·m/A times A divided by m is T. A current of a few amperes a couple of centimetres away gives a field comparable to the Earth's 25 to 65 μT, which is why Oersted saw a large swing rather than a twitch. OpenStax's own example needs about 25 A to make 1.0 × 10^-4 T at 5.0 cm, which agrees with the same formula.

The variant that flips the result. Put the compass on top of the wire instead of underneath. Directly above the wire the circling field points east, the opposite way, so the needle swings 68° east of north. Reverse the current with the compass underneath and it also swings east. Either change alone flips the deflection; make both and the needle swings west again. This is a quick test of whether you are using the right-hand rule rather than guessing.

Bottom line: a wire's field circles it, with B = μ₀I/(2πr). The right-hand rule fixes the direction at each point, and the field on one side of the wire is opposite to the field on the other.

Two wires: fields add, and wires push

Fields from several wires add as vectors. Take two long parallel wires 4.0 cm apart, each carrying 5.0 A. At the midpoint, 2.0 cm from each, each wire alone makes 50 μT. If the currents run in opposite directions, the right-hand rule shows that the two fields point the same way between the wires, and the midpoint field is 100 μT. If the currents run in the same direction, the two fields point opposite ways at the midpoint and cancel to zero. The arrangement of the currents, not just their size, decides the answer.

Each wire also sits in the field of the other, and a field pushes on a current, as the next section shows. The result, worked out with both right-hand rules, is a force on each wire per metre of length of

F/ℓ = μ0I1I2/(2πd)

where d is the separation. Currents in the same direction attract; currents in opposite directions repel. That is the opposite of the rule for charges, and it catches people out. For two wires 1.0 cm apart each carrying 10 A in the same direction, F/ℓ = (2.0 × 10^-7)(10)(10)/(0.010) = 2.0 × 10^-3 N on every metre, an attraction. Until 20 May 2019 this force defined the ampere: one ampere was the current that, in two long parallel wires 1 m apart in a vacuum, produced a force of exactly 2 × 10^-7 N per metre. NIST explains that the ampere is now based instead on the fixed value of the elementary charge.

The force on a current-carrying wire

Since a magnetic field pushes each moving charge sideways, it pushes on a wire that carries them. For a straight length ℓ of wire carrying current I in a uniform field B,

FB = IℓB sinθ

with θ the angle between the current and the field. Its direction follows right-hand rule 1, with the thumb along the current instead of the velocity of a positive charge: thumb along I, fingers along B, force out of the palm. A wire carrying 2.0 A through 5.0 cm of a 0.30 T field, perpendicular to it, feels F = (2.0 A)(0.050 m)(0.30 T) = 0.030 N; turn the wire so that the current runs along the field and the force vanishes. This sideways push on a current-carrying wire is what turns an electric motor: a loop of wire in a field is pushed one way on one side and the opposite way on the other, and turns.

Loops, coils and the electromagnet

Bend the wire into a circular loop and the circling fields of every part of the loop point the same way through the middle. At the centre the field points along the loop's axis, in the direction given by curling the fingers of your right hand round the loop in the direction of the current: your thumb points along the field. One side of the loop behaves as a north pole and the other as a south pole. A current loop is a magnetic dipole, which is the link back to Lesson 15: the atomic dipoles of a magnet are, in effect, tiny current loops.

Wind many loops into a coil and their fields add, producing a strong field along the coil's axis, stronger with more turns and with more current. Put an iron core inside, a nail will do, and the coil's field aligns the iron's domains, which add a much larger field of their own; OpenStax notes that adding ferromagnetic material produces greater field strengths. That is an electromagnet: a magnet you can switch on and off, and whose strength you can set with the current.

A home electromagnet has one hazard worth understanding, not just obeying. A coil of copper wire has very little resistance, so connected straight across a battery it draws a large current, limited mainly by the battery's own internal resistance, exactly as in Lesson 13. Three metres of thin copper wire 0.25 mm across has a resistance of about 1.1 Ω. On an AA cell with 0.29 Ω of internal resistance it draws about 1.1 A, turning about 1.3 W into heat in the coil and 0.4 W inside the cell. Thicker or shorter wire draws more current and heats faster. That is why the activity below connects the coil for a few seconds at a time and uses an ordinary alkaline cell, never a rechargeable lithium battery or a car battery.

Common misconceptions

• "A wire's magnetic field points along the wire, or straight out from it." It circles the wire, with no component along it or toward it.
• "Parallel currents repel, as like charges do." Currents in the same direction attract and opposite currents repel.
• "The field of a wire falls off as 1/r²." For a long straight wire it falls as 1/r: twice the distance, half the field.
• "A magnetic field pushes on any current-carrying wire." Not on a wire whose current runs along the field: F = IℓB sinθ is zero at 0° and 180°.
• "The iron in an electromagnet is the source of its magnetism." The current is the source; the iron multiplies the field by aligning its domains, and goes slack when the current stops.

Putting it together

Oersted's 1820 discovery that a current turns a compass needle has a simple quantitative form. A long straight wire carrying 5.0 A makes a field of B = μ₀I/(2πr) = 50 μT at 2.0 cm, circling the wire as right-hand rule 2 says. Against the Earth's 20 μT, that swings a compass beneath the wire 68° west of north, and 68° east if the compass is moved above the wire or the current reversed. The field falls as 1/r, and fields from several wires add as vectors: midway between equal opposite currents they double, between equal parallel currents they cancel. A field pushes on a current, F = IℓB sinθ: 0.030 N for 2.0 A in 5.0 cm of a 0.30 T field, and zero along the field. So parallel wires attract when their currents agree and repel when they oppose, with F/ℓ = μ₀I₁I₂/(2πd), the effect that defined the ampere until 2019. A current loop is a dipole with its field along its axis, a coil multiplies that field, and an iron core multiplies it again: that is all an electromagnet is. A low-resistance coil across a battery draws a large current and gets hot.

Sources

1. OpenStax. (2022). 22.9 Magnetic fields produced by currents: Ampere's law (Oersted, 1820; right-hand rule 2; Example 22.6). In College Physics 2e. openstax.org
2. OpenStax. (2022). 22.10 Magnetic force between two parallel conductors. In College Physics 2e. openstax.org
3. OpenStax. (2022). 22.7 Magnetic force on a current-carrying conductor; 22.8 Torque on a current loop: Motors and meters. In College Physics 2e. openstax.org
4. National Institute of Standards and Technology. (n.d.). SI redefinition: Ampere (the parallel-wire definition used until 20 May 2019, and the definition based on the elementary charge). nist.gov
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 12.3, Magnetism and current-carrying wires. apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Magnets and electromagnets (optional: vary the turns and current of a coil and watch a compass). phet.colorado.edu

A magnet that takes 23 seconds to fall 1.7 metres

Yan Levin, Fernando da Silveira and Felipe Rizzato are physicists at the Federal University of Rio Grande do Sul in Brazil. They held a copper pipe 1.7 m long upright and dropped a 6-gram cylindrical neodymium magnet down it. It took 22.9 seconds to come out of the bottom. An object with no magnetism, of the same mass and shape, covers the same distance in less than a second, as their 2006 paper in the American Journal of Physics notes. Free fall through 1.7 m takes 0.59 s, so the magnet was slowed by a factor of almost forty. It drifted down at a steady crawl of about 7 cm per second for nearly the whole way.

The explanation most people offer first is that the copper attracts the magnet and holds it back. It sounds reasonable, because magnets attract metals, and pipes are metal. This lesson tests that explanation against what actually happens and finds exactly where it fails. Then it replaces it with two laws, Faraday's and Lenz's, that also explain how every power station generator works.

Testing the obvious explanation

Three observations knock it down. First, hold the magnet against the outside of the copper pipe and let go: it does not stick. Copper is not ferromagnetic; as Lesson 15 showed, it is only very weakly diamagnetic. Second, hold the magnet still inside the pipe, and no force holds it up at all: the slowing appears only while the magnet is moving. Third, cut a long slit down the side of the pipe. The copper is still there, as close to the magnet as before, but the magnet now falls much faster. OpenStax describes the same effect with metal plates: slots limit the size of the current loops that can form in the metal, and the drag largely disappears.

So the slowing force needs two things that a simple attraction does not: a magnet in motion, and closed conducting paths running around the magnet. Both point to the same mechanism. A moving magnet changes the magnetic field passing through the rings of the pipe, and a changing field drives currents round any closed path it threads. Those currents have magnetic fields of their own, and it is their fields that push on the magnet. To make that precise, start with the quantity that changes.

Magnetic flux

The magnetic flux through a flat surface measures how much magnetic field passes through it. For a uniform field B and a flat area A,

ΦB = BA cosθ

where θ is the angle between the field and the area vector, an arrow perpendicular to the surface with length equal to its area. When the field passes straight through the surface, θ = 0 and the flux is BA, its largest value. When the field lies in the plane of the surface, θ = 90° and no field passes through, so the flux is zero. Flux is measured in tesla square metres, called webers (Wb). Its sign says whether the field passes through the surface in the same direction as the chosen area vector (positive) or the opposite direction (negative). For a closed surface, the area vector is taken to point outward. Flux can change in three ways: the field can change, the area can change, or the angle between them can change.

Faraday's law

Whenever the magnetic flux through a loop changes, an emf is induced around the loop, and its size is the rate of change of the flux:

|ε| = |ΔΦB/Δt|

This is Faraday's law. A coil of N turns has the same change of flux through each turn, so OpenStax writes its emf as N times as large. Note what Faraday's law does not say. A strong field induces nothing if it is steady; a large flux induces nothing if it is not changing. Only change counts, and the faster the change, the bigger the emf.

Givens. A single square loop of wire, 10 cm on a side, so A = 0.010 m², with a resistance of 0.10 Ω, sits with its plane perpendicular to a magnetic field. The field rises steadily from 0 to 0.50 T in 0.20 s.

The set-up in words. The loop lies flat on the face of a large electromagnet whose field points straight up through it while the current in the electromagnet is switched on.

Principle. Faraday's law for the change in flux, then Ohm's law for the induced current.

1. Change in flux, with θ = 0: ΔΦ = AΔB = (0.010 m2)(0.50 T - 0) = 0.0050 Wb.
2. Induced emf: |ε| = (0.0050 Wb)/(0.20 s) = 0.025 V.
3. Induced current: I = ε/R = (0.025 V)/(0.10 Ω) = 0.25 A, flowing only while the field is changing.

Sanity check. T·m²/s is a volt, since a tesla is N/(A·m): N·m/(A·s) = J/C = V. A few tens of millivolts from one loop is modest, which is why practical generators use many turns and fast changes; OpenStax's own single-loop example comes to 22.6 mV.

The variant that flips the result. Turn the loop so that its plane is parallel to the field lines and repeat. Now θ = 90°, and the flux is zero before, during and after the change. The emf is zero: the same rising field induces nothing in a loop it does not pass through. Or keep the loop flat and switch the field off over 0.20 s instead. The emf is the same size, 0.025 V, but the current runs the other way round the loop.

Lenz's law: the induced current opposes the change

Which way does the induced current flow? Lenz's law answers: the induced emf drives a current whose own magnetic field opposes the change in flux that produced it. If the flux through a loop is increasing, the induced current makes a field pointing against the external field inside the loop. If the flux is decreasing, the induced field points along the external field, trying to keep the flux up. To turn that into a direction round the loop, use the curled right hand from the last lesson. Point your thumb in the direction the induced field must point inside the loop, and your fingers curl in the direction of the induced current.

Push the north pole of a magnet toward a ring. The field through the ring points away from the north pole and is getting stronger, so the induced field must point back toward the magnet. The ring's near face becomes a north pole, and it repels the approaching magnet. Pull the magnet away and the flux falls; the induced current reverses, the near face becomes a south pole, and it attracts the retreating magnet. Either way the ring resists the motion. That is no accident. If the induced current helped the change instead, the magnet would be pulled in faster, which would induce a bigger current, which would pull harder, creating energy from nothing. Lenz's law is conservation of energy at work.

Why this matters: induction never gives something for nothing. Every induced current pushes back on whatever caused it, and the work done against that push is where the electrical energy comes from.

Back to the pipe: where the reasoning went wrong

Picture the pipe as a stack of copper rings. Below the falling magnet, the flux through each ring is increasing as the magnet approaches, so those rings carry currents that repel it from below. Above the magnet, the flux through each ring is decreasing as the magnet leaves, so those rings carry currents that attract it from above. Both push upward on the falling magnet, and the faster it falls, the faster the flux changes and the stronger the push. Within a short distance the upward push equals the magnet's weight, and it falls at a steady terminal speed. A magnet held still changes no flux, induces no current and feels no force, which is exactly what the second observation showed; and a slit breaks the rings, which is what the third showed.

The obvious explanation failed because it looked for a static attraction between materials, when the force depends on motion and on the pipe being able to carry current round the magnet. Energy confirms the new picture. At a steady speed of 1.7 m/22.9 s = 0.074 m/s the magnet's kinetic energy is not changing, so its gravitational potential energy must be going somewhere else. Levin and colleagues show it goes into ohmic heating of the pipe by the induced currents. The rate is mgv = (0.0060 kg)(9.8 m/s2)(0.074 m/s) = 4.4 × 10-3 W, a few milliwatts spread through 1.7 m of copper.

A rod on rails: motional emf

A simpler arrangement shows the whole energy chain in numbers. Two parallel metal rails, connected at one end through a resistor R, lie in a uniform field B perpendicular to the plane of the rails. A metal rod of length ℓ lies across them and is pushed along at speed v. The rod, the rails and the resistor form a loop whose area grows by ℓv every second, so the flux grows at Bℓv, and Faraday's law gives the motional emf

ε = Bℓv

With B = 0.40 T, ℓ = 0.50 m, v = 2.0 m/s and R = 2.0 Ω: ε = (0.40 T)(0.50 m)(2.0 m/s) = 0.40 V, and I = 0.40 V/2.0 Ω = 0.20 A. The field then pushes on the current in the rod with F = IℓB = (0.20 A)(0.50 m)(0.40 T) = 0.040 N, and by Lenz's law this force opposes the rod's motion. To keep the rod moving steadily you must push with 0.040 N, delivering Fv = 0.080 W, exactly the I²R = (0.20 A)²(2.0 Ω) = 0.080 W that heats the resistor. Push the rod the other way and the current reverses, and so does the drag, which still opposes you. Remove the resistor, leaving the circuit open, and the rod still has 0.40 V between its ends, but no current flows and there is no drag. An emf needs a closed path before it can drive a current, which is the slit pipe again.

A generator is the same idea turned into rotation. A coil spun in a magnetic field has a flux through it that rises and falls as it turns, so its emf alternates in direction, and the harder the circuit draws current, the harder the coil resists being turned. That resistance is what the turbine in a power station works against, and it is how mechanical work becomes electrical energy.

Common misconceptions

• "Copper attracts magnets." Copper is not ferromagnetic. The braking in the pipe comes from induced currents, and it vanishes when the magnet is still.
• "A strong magnetic field induces an emf." Only a changing flux does. A loop resting in a strong steady field has no emf.
• "The induced current always opposes the field." It opposes the change in flux. When the flux is decreasing, the induced field points the same way as the external field.
• "Without a complete circuit there is no emf." The emf is induced regardless; a closed path is needed for a current to flow.
• "A bigger flux means a bigger emf." The emf depends on how fast the flux changes, not on how large it is.

Pulling it together

A 6 g magnet takes 22.9 s to fall through 1.7 m of copper pipe. Copper does not attract it: it does not stick, it feels no force when still, and a slit pipe barely slows it. The magnet is slowed because its motion changes the flux through every ring of the pipe. Magnetic flux, Φ = BA cosθ in webers, is largest when the field passes straight through a surface and zero when it lies along it. Faraday's law says a changing flux induces an emf |ε| = |ΔΦ/Δt|. That is 0.025 V for 0.010 m² as the field rises 0.50 T in 0.20 s, and nothing if the loop is turned edge-on.

Lenz's law says the induced current opposes the change: a ring repels an approaching pole and attracts a retreating one. So the pipe pushes up on the magnet from both sides until it falls steadily, turning 4.4 mW of gravitational energy into heat. A rod on rails gives the motional emf ε = Bℓv, 0.40 V at 2.0 m/s in 0.40 T. The 0.040 N drag on it means the 0.080 W you supply is exactly the power heating the circuit: energy is conserved at every step.

Sources

1. Levin, Y., da Silveira, F. L., and Rizzato, F. B. (2006). Electromagnetic braking: A simple quantitative model. American Journal of Physics, 74, 815. https://doi.org/10.1119/1.2203645. Preprint: arxiv.org
2. OpenStax. (2022). 23.1 Induced emf and magnetic flux; 23.2 Faraday's law of induction: Lenz's law. In College Physics 2e. openstax.org
3. OpenStax. (2022). 23.3 Motional emf. In College Physics 2e. openstax.org
4. OpenStax. (2022). 23.4 Eddy currents and magnetic damping (slotted conductors). In College Physics 2e. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 12.4, Electromagnetic induction and Faraday's law. apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Faraday's law (optional: move a magnet through coils and watch the meter). phet.colorado.edu

Upside down at arm's length, right way up up close

Hold a polished soup spoon at arm's length and look into its bowl: your face appears small and upside down. Bring the spoon slowly toward your eye and, at some distance, the image blurs and vanishes; closer still, your face, or your eye, appears right way up and enlarged. Turn the spoon over and look into its back, and you see yourself small and upright at every distance. One curved piece of metal produces three different kinds of image, and the rules that decide which one you get are the subject of this lesson. They are worth learning as a procedure, because the exam asks you to run them both ways: by drawing rays and by calculating.

Light as rays, and the law of reflection

A light ray is a straight line drawn perpendicular to the wavefronts of a light wave, pointing the way the wave travels. Treating light as rays is the approach of geometric optics, and it works whenever the openings and objects light meets are large compared with its wavelength. It fails for the spreading of light through narrow openings, where the wave nature of light matters; that is the business of Module 6. A laser beam, a single narrow beam of one colour, is the nearest thing to an ideal ray, and a ray diagram shows the path of light before and after it meets matter.

When light strikes a surface, some of it is reflected. The law of reflection concerns the normal, the line perpendicular to the surface at that point. It says that the angle between the incoming ray and the normal equals the angle between the reflected ray and the normal: θi = θr. Both rays lie in the same plane as the normal, on opposite sides of it. A ray striking a mirror at 35° to the normal leaves at 35° to the normal.

Every surface obeys the law, but what you see depends on the surface's texture. On a smooth surface such as glass or polished metal, the normal points the same way over the whole area the light strikes, so a parallel beam is reflected as a parallel beam: specular reflection, which forms images. On a rough surface such as paper, the normal changes direction from point to point, so the reflected rays leave in many directions: diffuse reflection. That is why you can read a page from almost any angle but see your face only in a mirror.

Mirrors that converge, diverge, and do neither

A spherical mirror is a piece cut from the surface of a sphere. Its principal axis runs through the centre of the mirror and the centre of curvature C, the centre of the sphere it was cut from, a distance R, the radius of curvature, from the mirror. A concave mirror, curved inward like the bowl of a spoon, is a converging mirror: rays arriving parallel to the axis reflect toward a common point, the focal point F, in front of the mirror. A convex mirror, curved outward like the back of the spoon, is a diverging mirror: parallel rays reflect so that they appear to come from a focal point behind it. For a spherical mirror the focal point lies halfway between the mirror and its centre of curvature, so the focal length is

f = R/2

A plane mirror is the limiting case of a sphere of infinite radius: its focal point is infinitely far away, and it neither converges nor diverges light.

Real images and virtual images

A mirror forms an image of a point on an object when the rays leaving that point are redirected toward, or apparently from, a single point. If the reflected rays actually meet there, the image is real: it forms in front of the mirror, light really passes through it, and it can be projected onto a screen. If the reflected rays diverge, so that they only appear to have come from a point behind the mirror, the image is virtual. No light passes through it and no screen can catch it, though your eye can see it, as you see your face in a bathroom mirror.

The procedure, part one: three principal rays

To locate an image by drawing, trace two or three rays from the tip of the object whose paths you know without measuring any angles. For a concave mirror:

1. A ray parallel to the principal axis reflects through the focal point F.
2. A ray through F reflects parallel to the principal axis.
3. A ray striking the centre of the mirror, where the axis meets it, reflects at the same angle to the axis on the other side, as with a plane mirror.

For a convex mirror, the same rays apply with the focal point behind the mirror: a ray parallel to the axis reflects as if it came from F, and a ray aimed at F reflects parallel to the axis. The image of the tip is where the reflected rays cross, or where their backward extensions cross for a virtual image. Two rays fix it; the third is a check.

The procedure, part two: the mirror equation

The same answer comes from calculation. With s_o the object's distance from the mirror, s_i the image's distance and f the focal length,

1/si + 1/so = 1/f and |M| = |hi/ho| = |si/so|

where M is the magnification, the image height h_i divided by the object height h_o. The sign conventions carry the physics. The focal length is positive for a concave mirror and negative for a convex one. The image distance is positive for a real image, in front of the mirror, and negative for a virtual image, behind it. The exam's table quotes the magnification as a magnitude, so state the orientation in words: for a single mirror, a real image is inverted and a virtual image is upright. OpenStax writes the same relations with d_o and d_i and a signed magnification, m = -d_i/d_o, which is negative for an inverted image.

Givens. A concave mirror with radius of curvature 20 cm, so f = +10 cm. An object 2.0 cm tall stands upright on the axis, 30 cm in front of it.

The set-up in words. The mirror at the right of the page, its axis running left. C is 20 cm to its left, F is 10 cm to its left, and the object, a small arrow pointing up, is 30 cm to its left, beyond C.

Principle. The mirror equation for the position, the magnification for the size, and the sign of s_i for real or virtual; then a ray diagram as a check.

1. Position: 1/si = 1/f - 1/so = 1/(10 cm) - 1/(30 cm) = 2/(30 cm), so si = +15 cm.
2. Size: |M| = si/so = 15/30 = 0.50, so the image is 1.0 cm tall.
3. Description: s_i is positive, so the image is real, 15 cm in front of the mirror, between F and C. A real image from a single mirror is inverted. Real, inverted, reduced.

Sanity check. Draw it: the ray from the arrow's tip parallel to the axis reflects through F; the ray from the tip through F reflects parallel to the axis below it. They cross 15 cm from the mirror, below the axis: an inverted image half the object's height, as calculated. This is the spoon at arm's length.

The variant that flips the result. Move the object in to 5.0 cm, inside the focal point. Now 1/si = 1/10 - 1/5.0 = -1/10, so si = -10 cm: negative, a virtual image 10 cm behind the mirror, with |M| = 10/5.0 = 2.0, upright and twice as tall. In the ray diagram, the two reflected rays now spread apart in front of the mirror, and only their extensions behind it meet. The object crossed the focal point and the image switched from real and inverted to virtual and upright, which is what you see when the spoon comes close to your eye, and how a makeup mirror magnifies. With the object exactly at F, the reflected rays leave parallel and no image forms at all: that is where the spoon's image blurred and vanished.

A concave mirror of f = 10 cm, object by object:

So what?: where the object sits relative to F and C decides everything about the image. Find s_i from the mirror equation, read real or virtual from its sign, and size from |s_i/s_o|.

Convex and plane mirrors

A convex mirror always gives the same kind of image. With f = -20 cm and an object 60 cm away, 1/si = -1/20 - 1/60 = -4/60, so s_i = -15 cm and |M| = 0.25: virtual, upright, reduced, 15 cm behind the mirror. Wherever the object is, the image is smaller and upright, which is why OpenStax notes convex mirrors' use for security in shops: a small image covers a wide view. A plane mirror forms a virtual, upright image the same size as the object and exactly as far behind the mirror as the object is in front. Stand 1.5 m from a bathroom mirror and your image is 1.5 m behind the glass, 3.0 m from you.

Common misconceptions

• "A virtual image can be caught on a screen behind the mirror." No light reaches the place behind a mirror. A virtual image is only where the reflected rays appear to come from.
• "A concave mirror always magnifies." Only for objects inside its focal point. Beyond F its images are real and inverted, and beyond C they are reduced.
• "Your image in a plane mirror is on the surface of the glass." It is as far behind the mirror as you are in front of it.
• "The angle of reflection is measured from the mirror's surface." Both angles are measured from the normal.
• "Rough surfaces do not obey the law of reflection." Every tiny patch obeys it; the patches simply face different ways.

What you now know

A spoon shows your face small and inverted at arm's length, enlarged and upright up close, and small and upright from its back, and the rules behind those images are a procedure. Light travels as rays where wave effects can be ignored, and reflects with equal angles to the normal: specularly from smooth surfaces, which form images, and diffusely from rough ones. A concave mirror converges parallel rays to a focal point at f = R/2 in front of it. A convex mirror diverges them from a focal point behind, and a plane mirror does neither.

Three principal rays locate an image. The mirror equation 1/s_i + 1/s_o = 1/f with |M| = |s_i/s_o| calculates it, with f positive for concave mirrors and s_i positive for real images, which are inverted. An object 30 cm from a 10 cm concave mirror gives a real, inverted image half its size at 15 cm; at 5.0 cm the image flips to virtual, upright and doubled, 10 cm behind; at F there is no image. A convex mirror always gives virtual, upright, reduced images, a quarter size at 60 cm for f = -20 cm, and a plane mirror puts your image as far behind it as you stand in front.

Sources

1. OpenStax. (2022). 25.1 The ray aspect of light; 25.2 The law of reflection. In College Physics 2e. openstax.org
2. OpenStax. (2022). 25.7 Image formation by mirrors (ray-tracing rules; f = R/2; uses of convex and concave mirrors). In College Physics 2e. openstax.org
3. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 13.1 and 13.2 (plane and spherical mirrors only), and Table of Information: Equations. apcentral.collegeboard.org
4. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Geometric optics (optional: the current version includes mirrors as well as lenses). phet.colorado.edu

The coin that comes back

Put a coin in the bottom of an empty mug on a table. Lower your head until the near rim of the mug just hides the coin, then keep your head absolutely still while someone slowly pours water into the mug. Before the mug is half full, the coin comes back into view. Nothing moved: not the coin, not the mug, not your eye. Something happened to the light travelling from the coin to your eye, and it happened at the surface of the water.

That raises two questions this lesson answers. Why does light change direction when it crosses from water into air? And since the coin is now visible, where exactly does it appear to be? Answering them gives Snell's law, and Snell's law then predicts something stranger: light in water striking the surface steeply enough cannot get out at all.

Why light bends: a change of speed

Light travels through empty space at c = 299 792 458 m/s, a value fixed exactly by the definition of the metre, which the exam rounds to 3.00 × 10⁸ m/s. In a transparent material it travels more slowly, and the ratio is the material's index of refraction:

n = c/v

The larger the index, the slower the light. Some values from OpenStax's table, which vary slightly with the colour of the light:

In water, v = c/n = (3.00 × 10⁸ m/s)/1.333 = 2.25 × 10⁸ m/s, three quarters of its speed in a vacuum. Air is so close to 1 that it is treated as vacuum in almost every calculation.

A change of speed causes a change of direction whenever light meets a boundary at an angle. Picture a straight wavefront, a line of crests, arriving obliquely at the surface of water. One end of it enters the water first and slows down while the other end is still racing along in air, so the wavefront swings round, the way a cart veers when one wheel runs off the road into sand. The refraction of a ray, its change of direction on passing into a new medium, is that swing. A ray arriving along the normal, straight on, meets the boundary all at once and does not bend at all.

Snell's law

The amount of bending is fixed by Snell's law:

n1 sinθ1 = n2 sinθ2

where both angles are measured from the normal. Two rules follow, and the exam expects them as reasoning, not only as algebra. Passing into a medium of higher index, where light slows down, the ray bends toward the normal. Passing into a medium of lower index, where light speeds up, it bends away from the normal.

Worked problem: a laser beam into water, and back out

Givens. A narrow laser beam in air strikes a still water surface at 50° to the normal. Take n_air = 1.00 and n_water = 1.333.

The set-up in words. The beam slants down from the upper left onto a horizontal water surface; the normal is the vertical line through the point where it lands.

Principle. Snell's law at the air-water boundary.

1. sinθ2 = n1sinθ1/n2 = (1.00)(sin 50°)/1.333 = (0.766)/1.333 = 0.575.
2. θ2 = sin-1(0.575) = 35° from the vertical, inside the water.

Sanity check. Entering a slower medium, the beam should bend toward the normal, and 35° is smaller than 50°. Note what was divided: the sine of the angle, not the angle; 50°/1.333 = 38° would be wrong.

The variant that flips the result. Now send light the other way, upward from inside the water, striking the surface from below at 30° to the normal. Snell's law gives sinθ₂ = 1.333 sin 30° = 0.667, so θ₂ = 42°: the light emerges, bent away from the normal. Increase the angle in the water to 50° and the law demands sinθ₂ = 1.333 sin 50° = 1.02. No angle has a sine greater than 1, so there is no refracted ray at all. The surface reflects all of the light back into the water. The same boundary that let the beam in at 50° refuses to let it out at 50°.

Key idea: Snell's law relates sines, and it can fail to have a solution. When it does, the light does not leave the denser medium; it is totally reflected.

The coin explained: apparent depth

Back to the mug. Light from the coin travels up through the water and leaves at the surface into air, bending away from the normal on its way to your eye. Your eye and brain assume light travels in straight lines, so they trace each ray straight back along the direction it arrives from, to a point higher up and nearer to you than the coin really is. That is where you see the coin. With the water in place, the bent rays clear the rim of the mug, which is why the coin reappeared. Seen from directly above, the image sits at an apparent depth equal to the real depth divided by the index of the water: a coin under 8.0 cm of water appears to be 8.0/1.333 = 6.0 cm down, three quarters of its true depth. That follows from Snell's law for nearly vertical rays, where the sine and the tangent of a small angle are almost equal. It is also why a pond looks shallower than it is, and why a pencil standing in a glass of water seems bent at the surface.

Total internal reflection and the critical angle

The flipped variant above is an example of total internal reflection. It can happen only when light travels from a medium of higher index into one of lower index, and only beyond a particular angle. At the critical angle θ_c, the refracted ray just grazes along the surface, at 90° to the normal. Putting θ₂ = 90° into Snell's law gives

sinθc = n2/n1

For water into air, sinθ_c = 1.00/1.333 = 0.750 and θ_c = 48.6°; beyond that angle, all the light is reflected and none is transmitted. OpenStax gives 42.2° for polystyrene into air and only 24.4° for diamond, whose high index traps light so effectively that it escapes only at a few points, concentrated by the cut facets into the flashes that make diamonds sparkle.

The most important use is the optical fibre. A thin strand of glass, the core, is surrounded by a cladding of slightly lower index. Light entering the core strikes the boundary beyond the critical angle and is totally reflected, again and again, following the fibre even around gentle bends with very little loss. OpenStax lists the uses: telephone and internet signals, endoscopes that let doctors see inside the body, and fibres that deliver laser light for treatment.

Designing a measurement of n

In a school laboratory the classic measurement uses a semicircular block of clear plastic on a sheet of paper, with a ray box sending a narrow beam at the centre of its flat face. Varied: the angle of incidence θ₁, set with a protractor drawn on the paper. Measured: the angle of refraction θ₂ inside the block, found by marking where the beam leaves the curved face. The curved face is chosen because a ray travelling along a radius meets it along the normal and leaves without bending a second time. Analysis: plot sinθ₁ against sinθ₂; Snell's law predicts a straight line through the origin whose slope is n. A sample run with Plexiglas gave refraction angles of 7°, 13°, 19°, 25°, 31° and 35° for incidence angles of 10° to 60° in steps of 10°; the best line through the origin has a slope of 1.51, matching OpenStax's value. The same block, turned round so that the light starts inside the plastic, shows total internal reflection beyond a critical angle near sin^-1(1/1.51) = 41°.

Common misconceptions

• "Light bends because the second material is denser." What matters is the index of refraction, a measure of light's speed, not the mass density of the material.
• "Light always bends toward the normal at a water surface." Only when it enters the water. Leaving the water it bends away from the normal, or, beyond 48.6°, does not leave at all.
• "Total internal reflection can happen when light goes from air into glass." It needs light travelling toward a lower index. Light can always enter the slower medium.
• "The coin is closer than it looks." It is deeper than it looks: refraction raises the image to three quarters of the true depth.
• "Snell's law divides the angle by the index." It relates the sines of the angles; dividing angles gives wrong answers, badly so at large angles.

The takeaway

A coin hidden by a mug's rim reappears when water is poured in, because light leaving the water bends away from the normal and your eye traces it back to an image raised to three quarters of the true depth, 6.0 cm for 8.0 cm of water. Light slows in transparent materials, by a factor equal to the index of refraction, n = c/v, and a wavefront meeting a boundary at an angle swings round as one edge slows first. Snell's law, n₁ sinθ₁ = n₂ sinθ₂, bends light toward the normal entering a higher index (50° in air becomes 35° in water) and away from it entering a lower one, with no bending at normal incidence. Travelling toward a lower index beyond the critical angle, sinθ_c = n₂/n₁, light is totally reflected: 48.6° for water, 24.4° for diamond. Total internal reflection keeps light inside optical fibres and makes diamonds sparkle, and a plot of sinθ₁ against sinθ₂ measures n as its slope.

Sources

1. OpenStax. (2022). 25.3 The law of refraction (Table 25.1, indices of refraction; n = c/v). In College Physics 2e. openstax.org
2. OpenStax. (2022). 25.4 Total internal reflection (critical angles; optical fibres; diamonds). In College Physics 2e. openstax.org
3. National Institute of Standards and Technology. (2024). CODATA 2022 value: speed of light in vacuum (exact). NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 13.3, Refraction. apcentral.collegeboard.org
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Bending light (optional: measure angles and find the critical angle). phet.colorado.edu

One magnifying glass, three jobs

A magnifying glass with a focal length of 10 cm can do three different things depending only on where you hold it. Held a few centimetres above a page, it shows the print upright and several times larger. Held between a sunny window and a white wall across the room, about 10 cm from the wall, it paints a small, sharp, upside-down picture of the window on the wall. Held at just the right distance from a bright phone screen, it throws an enlarged, upside-down copy of the screen onto a card. The glass is the same each time. What changes is the object's distance compared with the focal length. This lesson is organised as a set of comparisons, because the exam's lens questions are almost always comparisons in disguise. They set converging against diverging, near the lens against far from it, and lens against mirror.

Comparison one: converging and diverging lenses

A thin lens bends light at its two surfaces by refraction. Its shape decides what it does to rays arriving parallel to its principal axis, the line through its centre perpendicular to its faces.

A lens has a focal point on each side, one focal length from its centre for a thin lens with the same medium on both sides. Read down the table and one fact stands out: a diverging lens on its own never forms a real image of a real object. So if a question describes an image on a screen made by a single lens, the lens must be converging.

The principal rays

Three rays from the tip of an object can be drawn without measuring an angle:

1. A ray parallel to the principal axis is refracted through the focal point on the far side of a converging lens, or away from the lens as if it came from the focal point on the near side of a diverging lens.
2. A ray through the centre of the lens passes straight through without changing direction, because the middle of a thin lens behaves like a thin sheet with parallel faces.
3. A ray through the near focal point of a converging lens, or heading toward the far focal point of a diverging lens, emerges parallel to the axis.

Where the emerging rays meet, on the far side, is a real image; where their backward extensions meet, on the object's side, is a virtual image.

The thin-lens equation, and a second comparison

The thin-lens equation has exactly the form of the mirror equation, with the same symbols on the exam's table:

1/si + 1/so = 1/f and |M| = |hi/ho| = |si/so|

The conventions: f is positive for a converging lens and negative for a diverging lens. The image distance s_i is positive for a real image, on the far side of the lens, and negative for a virtual image, on the same side as the object. For a single lens, a real image is inverted and a virtual image is upright. Here is the converging lens of f = 10 cm, object by object, with a concave mirror's behaviour for comparison:

Compare this with the concave mirror table in Lesson 19 and the pattern is identical. That is no coincidence, since both are converging devices obeying the same equation. The one difference is where the real images form. A lens passes light through, so its real images form on the far side; a mirror sends light back, so its real images form on the same side as the object. In both, real means the light really gets there and a screen can catch it.

Worked problem: the magnifying glass as a projector, then as a magnifier

Givens. A magnifying glass with f = +10 cm. A phone screen showing an arrow 3.0 cm tall is placed 15 cm from it.

The set-up in words. On a table, the phone stands upright with its screen toward the lens. The lens is held upright 15 cm away on the principal axis, and a white card on the far side can slide along the axis.

Principle. The thin-lens equation for the position, the magnification for the size, and the sign of s_i for real or virtual.

1. Position: 1/si = 1/f - 1/so = 1/(10 cm) - 1/(15 cm) = 1/(30 cm), so si = +30 cm.
2. Size: |M| = 30/15 = 2.0, so the image of the arrow is 6.0 cm tall.
3. Description: positive s_i, so the image is real, on the card 30 cm beyond the lens, inverted and doubled.

Sanity check. In the ray diagram, the ray from the arrow's tip parallel to the axis is bent down through F, 10 cm beyond the lens. The ray through the centre continues straight on, slanting down. They cross below the axis at 30 cm, giving an inverted image twice as tall. The object lies between f and 2f, the projector row of the table.

The variant that flips the result. Move the phone in to 8.0 cm, inside the focal length. Now 1/si = 1/10 - 1/8.0 = -1/40, so si = -40 cm and |M| = 40/8.0 = 5.0. The card catches nothing, wherever you slide it, because the rays leaving the lens spread apart. But look through the lens from the far side and you see the arrow upright, five times larger, apparently 40 cm away on the phone's side. The object crossed F and the image went from real, inverted and on the card to virtual, upright and inside the lens, the magnifying glass's normal job.

Remember: for a converging lens, the focal point is the dividing line. Beyond F, real and inverted images on the far side; inside F, virtual, upright, enlarged images seen through the lens.

Diverging lenses, briefly

A diverging lens of f = -15 cm viewing an object 30 cm away gives 1/si = -1/15 - 1/30 = -3/30, so s_i = -10 cm and |M| = 0.33. The image is virtual, upright and a third of the object's size, 10 cm from the lens on the object's side. Wherever the object is placed, the image stays virtual, upright and reduced, which is why a door peephole shows a small, wide view of the corridor.

Common misconceptions

• "A converging lens always magnifies." Only objects inside F give enlarged upright images; distant objects give small inverted real images, as in a camera.
• "A virtual image can be caught on a card." The rays never meet where the card is. The image exists only where the rays appear to come from, and the eye can see it through the lens.
• "Covering half a lens cuts off half the image." Every part of the lens receives light from every part of the object, so covering half of it leaves the whole image, only dimmer.
• "Real images form on the same side as the object, as with mirrors." For a lens, light passes through, so real images form on the far side.
• "The image of a distant object forms at 2f." It forms at almost exactly f, which is why imaging a window is the quickest way to measure a focal length.

Summing up

A 10 cm magnifying glass enlarges print held close and paints a small inverted window on a wall about 10 cm away. It also projects an inverted, doubled phone screen 30 cm away when the phone is 15 cm from it. One lens gives three images, decided by where the object sits relative to F. A converging lens, thicker in the middle with positive f, brings parallel rays to the focal point on its far side. A diverging lens, thinner in the middle with negative f, spreads them as if from the near focal point, and gives only virtual, upright, reduced images of real objects.

Three principal rays (parallel then through F, straight through the centre, through F then parallel) locate images. The thin-lens equation 1/s_i + 1/s_o = 1/f with |M| = |s_i/s_o| calculates them, with s_i positive for real images, which are inverted and on the far side. Moving the phone inside F, to 8.0 cm, turns the real image into a virtual one five times larger. Lens and mirror tables match line for line, except that a lens's real images form behind it and a mirror's in front.

Sources

1. OpenStax. (2022). 25.6 Image formation by lenses (ray-tracing rules; thin-lens equation; magnification). In College Physics 2e. openstax.org
2. OpenStax. (2022). 25.7 Image formation by mirrors (for comparison). In College Physics 2e. openstax.org
3. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 13.4, Images formed by lenses, and Table of Information: Equations. apcentral.collegeboard.org
4. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Geometric optics (optional: move an object past a lens's focal point and watch the image flip). phet.colorado.edu

A flick at one end of a string

Tie two metres of thin cotton string to two metres of heavy rope, stretch the pair taut across a room, and tie a scrap of ribbon to the middle of the string. Now give the free end of the string one sharp flick sideways. A hump runs along the string. When it passes the ribbon, the ribbon jumps sideways and drops back to exactly where it was. When the hump reaches the knot, two things happen at once: a smaller hump carries on into the rope, more slowly, and another comes back along the string toward your hand, upside down.

That one flick contains most of the ideas this lesson needs. Something travelled the length of the string while the string itself stayed where it was, so a wave carries energy and not matter. The hump travelled at a definite speed, set by the string, not by the flick. And at the knot, the change of material split the pulse in two and flipped one of the halves. This lesson follows the pulse through each of those events, then lets it repeat into a wave, and finally leaves strings behind for the waves that need no material at all.

What travels, and what does not

A wave pulse is a single disturbance that transfers energy from one place to another without transferring matter between them. A wave, in the sense this course uses, is a continuous, periodic disturbance with a well-defined wavelength and frequency, a pulse repeated over and over. The ribbon shows the distinction: each piece of string moves only up and down by a few centimetres, while the disturbance moves the whole length of the string.

In the string the disturbance is perpendicular to the direction the pulse travels, which makes it a transverse wave. In a longitudinal wave the disturbance is parallel to the direction of travel: push one end of a stretched spring toy forward and back, and a region of squeezed coils, followed by a region of spread coils, runs along it. Sound is a longitudinal wave in air. Its regions of high and low pressure are called compressions and rarefactions, and the amplitude of a sound wave can be described by the largest rise or fall in pressure above or below the undisturbed pressure. Waves on strings and sound are mechanical waves, which need a medium to travel through. Light and radio waves are electromagnetic and need none, which is why sunlight crosses empty space and sound does not.

What sets the speed

The speed of a wave depends on the kind of wave and on the medium, not on how hard or how often you flick. For a wave on a string under tension F_T, with mass m per length ℓ,

vstring = √(FT/(m/ℓ))

A tighter string carries waves faster, because a stretched piece pulls its neighbour back into line more strongly; a heavier string carries them more slowly, because each piece has more inertia to overcome. OpenStax's University Physics works the high E string of a guitar: 3.09 × 10^-4 kg per metre under 56.40 N gives 427 m/s.

Sound in air behaves the same way in principle: the speed is set by the air, and it rises with temperature. OpenStax gives v = (331 m/s)√(T/273 K), which is 331 m/s at 0 °C and 343 m/s at 20 °C, and lists sound speeds of 1480 m/s in fresh water and 5960 m/s in steel. All electromagnetic waves travel through a vacuum at the same speed, c = 3.00 × 10⁸ m/s.

The knot: part goes on, part comes back

Givens. The cotton string has a linear density of 2.0 g/m (0.0020 kg/m) and the rope 32 g/m (0.032 kg/m). The tension is 8.0 N throughout. A steady 10 Hz wave is sent along the string from your hand.

The set-up in words. String on the left, rope on the right, the knot between them, the wave arriving at the knot from the string side.

Principle. The wave speed in each part from v = √(F_T/μ), where μ = m/ℓ; the frequency fixed by the source; then λ = v/f; and the boundary rule for the reflected wave.

1. Speeds: string v1 = √[(8.0 N)/(0.0020 kg/m)] = √4000 m/s = 63 m/s; rope v2 = √[(8.0 N)/(0.032 kg/m)] = √250 m/s = 16 m/s.
2. Frequency: 10 Hz on both sides. The knot cannot oscillate at two frequencies at once, so the frequency of a wave never changes when it crosses a boundary.
3. Wavelengths: λ1 = v1/f = (63 m/s)/(10 Hz) = 6.3 m in the string and λ2 = (16 m/s)/(10 Hz) = 1.6 m in the rope.
4. At the knot, part of the wave is transmitted into the rope and part is reflected back along the string, both with smaller amplitudes than the incoming wave. Because the transmitted wave travels into a medium where it slows down, the reflected wave is inverted: a crest comes back as a trough.

Sanity check. The rope is 16 times heavier per metre, so the speed should be √16 = 4 times smaller, and 63/16 = 4. The wave is the same 10 Hz on both sides, so a slower wave must be a shorter one. And the inverted reflection matches the limiting case: a knot tied to something infinitely heavy is a fixed end, and reflection at a fixed end is inverted, as OpenStax describes.

The variant that flips the result. Send the wave the other way, from the rope into the string. Now the transmitted wave speeds up, and the reflected wave comes back upright, a crest returning as a crest, like a reflection from a free end. The frequencies and the two speeds are the same as before; only the direction of the crossing has changed, and with it the sign of the reflection.

Worth holding on to: the medium sets the speed, the source sets the frequency, and the wavelength is whatever fits, λ = v/f. At a boundary the frequency is unchanged, and a reflection into a slower medium is inverted while a reflection into a faster one is not.

From pulse to wave: describing a periodic wave

Keep flicking regularly and the string carries a periodic wave. Its period T is the time for one complete oscillation, its frequency f is the number of oscillations per second, and T = 1/f. Its wavelength λ is the distance between successive corresponding points, crest to crest or trough to trough, and its amplitude A is the greatest displacement from equilibrium. Amplitude and frequency are independent: you can make a tall slow wave or a small fast one on the same string. The energy a wave carries increases with its amplitude, and for a given amplitude it also increases with frequency. For sound, a larger amplitude is heard as louder and a higher frequency as a higher pitch; OpenStax gives the range of human hearing as about 20 to 20 000 Hz, so a 440 Hz tone in air at 20 °C has a wavelength of (343 m/s)/(440 Hz) = 0.78 m.

A sinusoidal wave can be described two ways, and the exam uses both. Watching one point of the string as time passes gives its displacement as a function of time, y(t) = A cos(2πft), a graph whose horizontal axis is time and whose repeat length is the period. Freezing the whole string at one instant gives the displacement as a function of position, y(x) = A cos(2πx/λ), a snapshot whose repeat length is the wavelength. For a wave of amplitude 2.0 cm, frequency 5.0 Hz and wavelength 0.80 m, a snapshot at one instant reads:

and the same point watched over time repeats every T = 1/(5.0 Hz) = 0.20 s. The two graphs look alike but answer different questions; the speed connects them, v = λ/T = fλ = (5.0 Hz)(0.80 m) = 4.0 m/s.

Polarization: a test only transverse waves pass

Thread a rope through a picket fence and shake it. Shake it up and down, parallel to the slats, and the wave passes; shake it side to side, across the slats, and the fence stops it. A transverse wave can oscillate in any direction perpendicular to its travel, and a wave that oscillates in only one such direction is polarized. A longitudinal wave oscillates along its direction of travel, so there is nothing for a fence to select: sound cannot be polarized.

Light is transverse, and its direction of polarization is the direction of its electric field. Ordinary light from the sun or a bulb is unpolarized, a jumble of all directions. A polarizing filter passes only the part oscillating along its axis, which, as OpenStax's University Physics puts it, decreases the intensity of unpolarized light by a factor of 2, where intensity is the average power carried per unit area. A second filter turned at 90° to the first passes nothing at all. Light reflected from a horizontal surface, such as a wet road, is partly polarized horizontally, which is why sunglasses with vertical polarizing axes cut glare. The screens of many devices emit polarized light too, since liquid-crystal displays work by rotating the polarization of light.

Electromagnetic waves and the spectrum

An electromagnetic wave consists of oscillating electric and magnetic fields, perpendicular to each other and to the direction the wave travels, so it is transverse. It needs no medium, and in a vacuum every electromagnetic wave travels at c. Electromagnetic waves are sorted into categories by wavelength, which in order of decreasing wavelength, from kilometres down to picometres, run: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. The visible band is narrow, roughly 400 to 750 nm according to OpenStax, and within it the colours run, again in order of decreasing wavelength, red, orange, yellow, green, blue and violet. The exam expects the order, not the exact boundaries, which in any case overlap.

The wave relation holds for light as for strings. A green laser pointer at 532 nm has a frequency of f = c/λ = (3.00 × 10⁸ m/s)/(532 × 10^-9 m) = 5.64 × 10¹⁴ Hz. Shone into water, its frequency stays the same, as at the knot, and its wavelength shrinks to 532 nm/1.333 = 399 nm, because light travels more slowly there.

Common misconceptions

• "A wave carries the material along with it." It carries energy. The ribbon on the string goes up and down and stays where it was.
• "Flicking harder makes the pulse travel faster." A bigger flick makes a bigger amplitude. The speed is set by the tension and the mass per length.
• "A wave changes frequency when it enters a new medium." The frequency is set by the source and stays fixed; the speed and the wavelength change.
• "Every reflected pulse is inverted." Only a reflection into a slower medium, or at a fixed end. Into a faster medium, or at a free end, the reflection is upright.
• "Sound can be polarized like light." Sound is longitudinal, and only transverse waves can be polarized.
• "Light needs air to travel through." Electromagnetic waves need no medium; sunlight crosses the vacuum of space, and sound cannot.

Looking back

One flick on a string tied to a rope showed a wave carrying energy without carrying the string, travelling at a speed set by the medium, and splitting at the knot into a transmitted pulse and a reflected one. On a string, v = √(F_T/(m/ℓ)): 63 m/s for 2.0 g/m at 8.0 N and 16 m/s for the 32 g/m rope; sound in air goes 331 m/s at 0 °C and 343 m/s at 20 °C; light in a vacuum goes at c. At a boundary the frequency stays fixed and the wavelength changes, from 6.3 m to 1.6 m for a 10 Hz wave entering the rope, and the reflection is inverted going into a slower medium and upright going into a faster one. Periodic waves have T = 1/f and λ = v/f; amplitude is independent of frequency, energy grows with both, and the wave can be graphed against time or against position. Transverse waves can be polarized and longitudinal ones cannot: a filter halves the intensity of unpolarized light, crossed filters block it, and sunglasses cut horizontally polarized glare. Electromagnetic waves are transverse, need no medium, and run from radio to gamma rays, with visible light, red to violet, in a narrow band near 400 to 750 nm.

Sources

1. OpenStax. (2016). 16.3 Wave speed on a stretched string (Example 16.5, guitar strings); 16.5 Interference of waves (reflection at fixed and free ends and at a junction of strings). In University Physics Volume 1. openstax.org
2. OpenStax. (2022). 17.2 Speed of sound, frequency, and wavelength (speed against temperature; speeds in various media; range of hearing). In College Physics 2e. openstax.org
3. OpenStax. (2022). 16.9 Waves; 24.3 The electromagnetic spectrum. In College Physics 2e. openstax.org
4. OpenStax. (2016). 1.7 Polarization. In University Physics Volume 3. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 14.1 to 14.4 (ordering of the spectrum required, exact wavelength ranges not). apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Wave on a string (optional: change tension and density, and compare fixed and loose ends). phet.colorado.edu

Two phones, one throbbing note

Put two phones side by side, each running a tone-generator app, one set to 440 Hz and the other to 443 Hz, both at a comfortable volume. You do not hear two notes. You hear a single note that swells and fades, loud, soft, loud, soft, exactly three times every second. Set both to 440 Hz and the throbbing stops; set the second to 446 Hz and it speeds up to six times a second. Where does a rhythm of three per second come from, when neither phone is producing anything at 3 Hz?

The answer is the principle that also explains why a guitar string plays only certain notes and why a closed pipe sounds different from an open one: waves that meet simply add. This lesson starts from that principle, uses it on the phones, then on waves trapped between boundaries, and ends with the change of pitch of a passing siren.

Waves pass through each other: superposition

Send two pulses toward each other along a stretched string, one up and one down. They do not bounce off each other. While they overlap, the displacement of each point of the string is the sum of the displacements the two pulses would produce separately; this is the principle of superposition. Then they pass through each other and carry on, each exactly as it was before. If the two displacements are in the same direction they add to something larger, which is constructive interference; if they are in opposite directions they partly or completely cancel, which is destructive interference. Two equal and opposite pulses that overlap exactly leave the string flat for an instant, though the pulses are not destroyed: the string is still moving, and a moment later both pulses re-emerge.

The phones explained: beats

Givens. Two steady sound waves of equal amplitude, at f₁ = 440 Hz and f₂ = 443 Hz, reach your ear together.

The set-up in words. The two phones sit side by side about a metre in front of you, so the two waves arrive along the same path.

Principle. Superposition. At some moments the two waves arrive in step, crest with crest, and add; a short time later one has gained half a cycle on the other, they arrive crest with trough, and cancel. The loudness therefore rises and falls, once for every cycle that one wave gains on the other.

1. In one second the 443 Hz wave completes 443 cycles and the 440 Hz wave 440, so the faster one gains 3 cycles on the slower one each second.
2. Each cycle gained brings one loud moment and one quiet moment, so the loudness swells three times a second: fbeat = |f1 - f2| = |440 Hz - 443 Hz| = 3 Hz.
3. The note you hear has a pitch close to the average of the two, 441.5 Hz.

Sanity check. Equal frequencies give |f₁ - f₂| = 0, no beats, as observed; 440 Hz and 446 Hz give 6 beats per second. The absolute value matters: 437 Hz against 440 Hz also gives 3 beats per second.

The variant that flips the result. That absolute value is a genuine ambiguity, and a musician tuning a string meets it constantly: 3 beats per second against a 440 Hz reference means the string is at either 437 Hz or 443 Hz. OpenStax notes that piano tuners listen for beats and adjust the string until they go away. To decide which way to turn the peg, raise the string's pitch slightly and listen. If the beats slow down, the gap is shrinking, so the string was low, at 437 Hz, and is approaching 440 Hz. If they speed up, it was high, at 443 Hz. The same small adjustment gives opposite instructions depending on which side of 440 Hz you started.

What matters here: waves add point by point and pass through each other unchanged. Two close frequencies drift in and out of step, and the beat frequency is their difference.

Waves trapped between boundaries: standing waves

Shake one end of a string whose other end is tied to a wall, and the wave reflects from the wall and travels back through the incoming wave. The two waves, moving in opposite directions, superpose. At most frequencies the result is a mess, but at certain frequencies it settles into a standing wave, a pattern that oscillates in place without travelling. Some points never move at all, the nodes, where the two waves always cancel; halfway between them are the antinodes, where the amplitude is largest. The distance from one node to the next is half a wavelength.

Which wavelengths can fit is decided by the size of the region and by what happens at its ends. A string fixed at both ends must have a node at each end, so it must hold a whole number of half wavelengths: L = nλ/2, giving λ_n = 2L/n and f_n = nv/(2L), for n = 1, 2, 3, .... The longest wavelength, n = 1, is the fundamental or first harmonic; the next is the second harmonic, and so on.

Givens. OpenStax's University Physics gives the speed of waves on a guitar's high E string as 427 m/s. Take the vibrating length of the string as 0.650 m.

The set-up in words. The string is pinned at the nut at one end and the bridge at the other, both nodes.

Principle. λ_n = 2L/n for a string fixed at both ends, and f = v/λ.

1. Fundamental: λ1 = 2L = 1.30 m, so f1 = (427 m/s)/(1.30 m) = 328 Hz.
2. Second harmonic: λ₂ = L = 0.650 m and f₂ = 657 Hz; third: λ₃ = 0.433 m and f₃ = 985 Hz. Each is a whole-number multiple of f₁.

Sanity check. Pressing the string halfway along the fingerboard halves L and should double the frequency, which is exactly the octave a guitarist hears at the twelfth fret.

Standing waves in pipes

Sound in a pipe does the same thing, with one change in the rule for the ends. At a closed end the air cannot move, so it is a node; at an open end the air moves most freely, so it is an antinode. A pipe open at both ends, like one closed at both, holds a whole number of half wavelengths; but a pipe closed at one end has a node at one end and an antinode at the other, so it holds an odd number of quarter wavelengths, and only the odd harmonics can form.

The flip is in the table. Close one end of an open 0.25 m pipe and its fundamental drops an octave, from 686 Hz to 343 Hz, because the pipe now holds a quarter wavelength instead of a half, and every even harmonic disappears. The two pipes, the same length, sound different notes with different tone qualities. A string with one end fixed and the other free to slide on a ring would behave like the closed pipe, with a node at one end and an antinode at the other.

Measuring the speed of sound in a glass of water

A tall glass or a measuring cylinder partly filled with water is a pipe closed at one end, the water surface, with an adjustable length of air above it. Hold a phone playing a steady 1000 Hz tone just above the open top and slowly change the water level, or slide a tube up and down in a tall jug of water, listening for the air-column lengths at which the sound becomes suddenly loud. Those are resonances: lengths at which the column holds a standing wave of the phone's frequency. In a sample run the first resonance came at 7.3 cm of air and the second at 24.5 cm. Successive resonances in a closed pipe differ by half a wavelength, so λ = 2(24.5 cm - 7.3 cm) = 34.4 cm and v = fλ = (1000 Hz)(0.344 m) = 344 m/s. Using the difference between two lengths is the key design choice: the antinode at an open end actually sits a little beyond the rim, so a single length gives the quarter wavelength only approximately, while the difference cancels that uncertainty. The first length, 7.3 cm, is indeed a little less than a quarter of 34.4 cm, which is 8.6 cm.

The Doppler effect

A siren on a passing ambulance sounds higher as it approaches and lower as it moves away; OpenStax describes the high-pitched scream of a motorcycle shifting to a lower roar as it passes. This is the Doppler effect. A moving source emits each wavefront from a little further along its path than the last, so the wavefronts are bunched together ahead of it and spread apart behind it. An observer ahead receives wavefronts more often, a higher frequency; an observer behind receives them less often, a lower frequency. The same happens if the observer moves toward or away from a stationary source. The source itself never changes its frequency, the rest frequency; what changes is the frequency observed.

The College Board asks for this only qualitatively, and the rules are three. A source and observer moving with the same velocity have no relative motion, so the observed frequency equals the rest frequency. Approaching gives a higher observed frequency, receding a lower one. And the greater the relative velocity, the greater the difference between the observed and rest frequencies.

Common misconceptions

• "Colliding pulses bounce off each other." They pass through each other; while they overlap, their displacements add.
• "Beats are a third sound at 3 Hz." They are a slow rise and fall in the loudness of one note, caused by two waves drifting in and out of step.
• "A pipe closed at one end has all the harmonics." A node at one end and an antinode at the other allow only odd harmonics.
• "A standing wave carries energy along the string." The pattern stays in place; energy sloshes between nodes but does not travel along.
• "The Doppler effect changes the frequency the source emits." The source is unchanged. The observed frequency changes because of the relative motion.

Recap

Two phones at 440 Hz and 443 Hz produce one note that throbs three times a second, because waves superpose: they add point by point, pass through each other unchanged, and two close frequencies drift in and out of step to give beats at f_beat = |f₁ - f₂|, which can be 3 Hz from either side, an ambiguity a tuner resolves by nudging the pitch and listening. Waves confined between boundaries form standing waves, with nodes where they always cancel and antinodes where they are largest. A string fixed at both ends holds whole numbers of half wavelengths, so a 0.650 m string with waves at 427 m/s sounds 328 Hz and all its multiples; an open pipe does the same, 686 Hz for 0.25 m, while a pipe closed at one end holds odd numbers of quarter wavelengths, 343 Hz and only odd harmonics. Two resonances 17.2 cm apart at 1000 Hz in a glass of water give the speed of sound, 344 m/s. A source and an observer approaching each other produce a higher observed frequency, receding a lower one, more so the faster they move relative to each other, and no shift at all when they share the same velocity.

Sources

1. OpenStax. (2022). 16.10 Superposition and interference (beats; standing waves on a string). In College Physics 2e. openstax.org
2. OpenStax. (2022). 17.5 Sound interference and resonance: Standing waves in air columns. In College Physics 2e. openstax.org
3. OpenStax. (2022). 17.4 Doppler effect and sonic booms. In College Physics 2e. openstax.org
4. OpenStax. (2016). 16.3 Wave speed on a stretched string (the high E string, 427 m/s). In University Physics Volume 1. openstax.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 14.5 (Doppler effect, qualitative only) and 14.6. apcentral.collegeboard.org

A rainbow on the back of a disc

Tilt a CD under a lamp and its silver underside breaks the white light into bands of colour. Nothing on the disc is coloured. What it has is a single spiral track. The industry standard for the CD, ECMA-130, specifies the distance between neighbouring turns of that spiral, the track pitch, as 1.6 µm, within 0.1 µm. That is only about two and a half times the wavelength of red light. When light meets a regular pattern on a scale that close to its own wavelength, it stops behaving like a stream of rays and starts behaving unmistakably like a wave.

This lesson builds the wave optics of openings in a fixed order: one opening, then two, then many, which is what a CD is. Each step is a calculation you can carry out with a ruler. The last one measures the CD's track pitch with a laser pointer and checks it against the standard.

Diffraction: waves spread through openings

Diffraction is the spreading of a wave around the edges of an obstacle or through an opening. It is strongest when the opening is comparable in size to the wavelength. Sound, with wavelengths around a metre, bends easily round a doorway. Light, with wavelengths below a micrometre, passes through the same doorway in a sharp-edged beam, which is why ray optics worked so well in Module 5. Send light through a slit a fraction of a millimetre wide, though, and it fans out, and on a distant screen it forms a pattern of bright and dark bands.

This pattern from a single opening is called single-slit diffraction. The bands come from interference among the wavelets leaving different parts of the opening, which travel different distances to each point on the screen. The amount of interference between two wavefronts depends on their path length difference, ΔD. For a single slit of width a, the screen is dark where light from the top edge and the bottom edge of the slit differs in path by a whole number of wavelengths. The slit can then be divided into pairs of strips whose wavelets cancel:

ΔD = a sinθ = mλ, for dark fringes, m = 1, 2, 3, ...

For small angles, which the exam allows for single and double slits, sinθ is close to y/L. Here y is the distance on the screen from the centre of the pattern and L is the distance from slit to screen, so a(ymin/L) ≈ mλ. Red light of 650 nm through a slit 0.10 mm wide onto a screen 2.0 m away gives its first dark fringes at y = λL/a = (650 × 10^-9 m)(2.0 m)/(1.0 × 10^-4 m) = 13 mm on either side of the centre. So the central bright band is 26 mm wide, 260 times the width of the slit. Make the slit narrower and the band gets wider: the smaller the opening, the more the light spreads. The pattern also depends on the shape of the opening; a pinhole gives a bright disc surrounded by faint rings.

Two slits: the experiment that settled the argument

OpenStax dates Thomas Young's double-slit experiment to 1801. Light from one source passes through two narrow slits a distance d apart and falls on a screen, which shows a row of evenly spaced bright and dark fringes. Particles flying in straight lines would light two strips; only waves, adding where they arrive in step and cancelling where they arrive out of step, produce a striped pattern. These fringes of double-slit interference were the evidence that light is a wave. The path difference from the two slits to a point on the screen is ΔD = d sinθ. The bright fringes are where it is a whole number of wavelengths, and the dark ones where it is a whole number plus a half:

d sinθ = mλ (bright), and d sinθ = (m + ½)λ (dark), with m = 0, 1, 2, ...

Givens. Red laser light of wavelength 650 nm falls on two slits 0.25 mm apart. The screen is 2.0 m away.

The set-up in words. A laser on a bench shines through a slide carrying two parallel slits. The light spreads and falls on a white wall 2.0 m beyond, where a row of red spots, the fringes, runs across the wall perpendicular to the slits.

Principle. d sinθ = mλ for bright fringes, with the small-angle form d(y_max/L) ≈ mλ.

1. Position of the m-th bright fringe: ym ≈ mλL/d.
2. Spacing between neighbouring bright fringes: Δy = λL/d = (650 × 10-9 m)(2.0 m)/(0.25 × 10-3 m) = 5.2 × 10-3 m, 5.2 mm.
3. The third bright fringe from the centre is at 3 × 5.2 mm = 15.6 mm, an angle of only 0.0078 rad, under half a degree.

Sanity check. The angles are tiny, so the small-angle approximation is excellent. A wavelength of a fraction of a micrometre and a slit separation of a quarter of a millimetre give fringes a few millimetres apart on a wall two metres away. That is large enough to measure with a ruler, which is how Young's kind of experiment measures wavelengths.

The variant that flips the result. Halve the slit separation to 0.125 mm and the fringes move apart, to 10.4 mm: bringing the slits closer spreads the pattern. Switch to blue light at 450 nm and they close up, to 3.6 mm. In wave optics, smaller structures and longer wavelengths both mean wider patterns. Real double slits also have a width, so the evenly spaced fringes appear within the broader envelope of each slit's single-slit pattern, bright in the middle and fading at the sides.

The upshot: fringe positions come from path differences. Whole wavelengths give bright fringes, half-odd ones give dark fringes, and the pattern's scale is λL divided by the size of the openings or their spacing.

Many slits: gratings, and a CD

A diffraction grating is a large number of evenly spaced parallel slits, or reflecting lines. Its bright maxima obey the same equation as two slits, d sinθ = mλ. But with so many slits the maxima are much narrower and brighter, and the spaces between them are darker, which is what makes gratings so useful for analysing light. OpenStax notes that a grating works by reflection as well as transmission, as in a CD. Because grating lines are usually only a micrometre or two apart, the angles are large, and the small-angle approximation no longer works. Find the angle from the measured distances with tanθ = y/L, then use sinθ in the grating equation.

White light shows why gratings make rainbows. The central maximum, m = 0, is white, since every wavelength satisfies it at θ = 0. In every other order each wavelength lands at its own angle, and longer wavelengths at larger angles, so each order is a spectrum with violet nearest the centre and red farthest out.

Givens. A red laser pointer of wavelength 650 nm shines perpendicularly onto a CD, whose track pitch, from ECMA-130, is 1.6 µm. The reflected spots fall on a card 20.0 cm from the disc.

The set-up in words. The laser passes through a small hole in a white card and strikes the disc squarely. The straight-back reflection, m = 0, returns through the hole; the diffracted spots land on the card on either side of it.

Principle. d sinθ = mλ, with sinθ no larger than 1, and tanθ = y/L for the positions on the card.

1. First order: sinθ1 = λ/d = (650 nm)/(1600 nm) = 0.406, so θ₁ = 24.0° and the spot sits y = L tanθ = (20.0 cm)(0.445) = 8.9 cm from the hole.
2. Second order: sinθ₂ = 2(0.406) = 0.813, so θ₂ = 54.3° and y = (20.0 cm)(1.39) = 27.9 cm, off the edge of most cards.
3. Third order: sinθ₃ would be 1.22, which is impossible: there is no third order.

Sanity check. The small-angle shortcut, y ≈ mLλ/d, would put the first spot at 8.1 cm instead of 8.9 cm, a 9 percent error, and the second at 16 cm instead of 28 cm, a disaster. At grating angles the shortcut must be dropped. The orders really do run out, because d is only 2.5 wavelengths.

The variant that flips the result. Replace the CD with a DVD, whose track pitch ECMA-267 specifies as 0.74 µm. The first order now needs sinθ₁ = 650/740 = 0.878, so it appears at a steep 61°, and the second order would need sinθ₂ = 1.76, which is impossible. The CD shows two orders on each side; the DVD, with its finer tracks, shows only one, thrown out at a much larger angle. A green 532 nm pointer on the DVD gives its single first order at 46°.

Designing the measurement safely

Run backwards, the same geometry measures the track pitch. Measure the distance L from the disc to the card and the distance y from the hole to a first-order spot. Find θ from tanθ = y/L, and calculate d = λ/sinθ, using the wavelength printed on the laser's label. In a sample run with L = 10.0 cm, the two first-order spots were 8.8 cm apart, so y = 4.4 cm, tanθ = 0.44 and θ = 23.7°. Then d = 650 nm/sin 23.7° = 1610 nm, 1.61 µm, inside the ECMA-130 tolerance of 1.5 to 1.7 µm. Measuring the distance between the two first-order spots and halving it removes any error in locating the centre, and repeating at two or three distances L checks the result.

The laser must be treated with care. Use only a low-power pointer labelled Class 2 (Class II under the FDA's labelling), under the supervision of a teacher or other adult. The FDA advises against giving laser pointers to children. The FDA's rules are plain: never aim or shine a laser pointer at anyone, and never look directly into any laser beam. Because the CD is itself a mirror, set the geometry up before switching on, so that every reflected beam lands on the card. Keep the whole arrangement below eye level, and never point the laser at any other shiny surface.

Common misconceptions

• "A narrower slit makes a narrower pattern." It makes a wider one: the central band's half-width is λL/a.
• "The CD's colours come from dyes or a coating." They come from interference among reflections from evenly spaced tracks; each colour leaves at its own angle.
• "The small-angle formula always works." It is fine for double slits a fraction of a millimetre apart, but badly wrong at the 24° and 54° angles of a CD.
• "In a grating spectrum, violet is deviated most." Red, with the longest wavelength, is deviated most; the central maximum stays white.
• "Dark fringes mean light was destroyed." The energy is redistributed: missing from the dark fringes, it is concentrated in the bright ones.

What to carry forward

A CD's single spiral track, 1.6 µm from turn to turn under the ECMA-130 standard, splits white light into colours. Light spreads and interferes when it meets structures close to its wavelength. A single slit of width a gives dark fringes at a sinθ = mλ. So 650 nm light through 0.10 mm makes a central band 26 mm wide at 2.0 m, wider for narrower slits. Two slits d apart give bright fringes at d sinθ = mλ and dark ones halfway between, evenly spaced by λL/d: 5.2 mm for 650 nm, 0.25 mm and 2.0 m. It was this striped pattern that let Young show in 1801 that light is a wave.

A grating obeys the same equation with sharper, brighter maxima at large angles, where tanθ = y/L must replace the small-angle shortcut. White light gives a white centre and spectra with red farthest out. A red pointer on a CD gives orders at 24° and 54° and no third; on a 0.74 µm DVD only one order, at 61°. Measuring y and L for the first order measures the track pitch, 1.61 µm in a sample run. The laser is a Class 2 pointer, never aimed at anyone or looked into.

Sources

1. Ecma International. (1996). Standard ECMA-130: Data interchange on read-only 120 mm optical data disks (CD-ROM) (2nd ed.), clause 11.3, track pitch 1.6 µm ± 0.1 µm. ecma-international.org
2. Ecma International. (2001). Standard ECMA-267: 120 mm DVD, read-only disk (3rd ed.), track pitch 0.74 µm ± 0.03 µm. ecma-international.org
3. OpenStax. (2022). 27.3 Young's double slit experiment; 27.4 Multiple slit diffraction; 27.5 Single slit diffraction. In College Physics 2e. openstax.org
4. U.S. Food and Drug Administration. (n.d.). Illuminating facts about laser pointers; and Frequently asked questions about lasers. fda.gov
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 14.7 and 14.8 (small-angle approximation valid for single and double slits). apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Wave interference (optional: slits of adjustable width and separation). phet.colorado.edu

A film that goes black

Dip a wire loop in soapy water and hold it upright in front of a window. The film shimmers with horizontal bands of colour, which slowly slide downward as the liquid drains. Then the top of the film turns black, a dark band that grows, and a few seconds later the film bursts. The black is not a hole: the film is still there, holding together, and it reflects nothing.

The explanation most people give is that the film at the top has become too thin to reflect any light. It sounds sensible, because less soap seems as if it ought to mean less reflection. This lesson tests that idea and finds the exact point where it goes wrong. It replaces it with the two-reflection model. That model also explains the colours above the black band, the rainbow swirls on a bubble, and the coatings that stop camera lenses and spectacles from reflecting.

Testing the explanation

Start with what reflection needs. When light meets a boundary between two transparent media, some of it is reflected, some is transmitted, and, in real materials, a little is absorbed. Reflection happens at the boundary itself, because of the change of index, however thin the layer beyond it is. A soap film has two boundaries, air to soapy water at the front and soapy water to air at the back, and each of them reflects a few percent of the light. A thinner film still has both. So the "too thin to reflect" explanation must be wrong about something: a very thin film does produce two reflected waves. They must be disappearing because they cancel each other, and to see why they cancel, we need to compare them.

Two reflections and a half-cycle flip

The two reflected waves differ in two ways. The first is a rule about reflection itself, which OpenStax states this way. A wave reflecting from a medium of higher index of refraction than the one it is travelling in undergoes a phase change of half a wavelength, 180°. A crest comes back as a trough, just as a pulse on a light string reflects inverted from a heavy rope. Reflection from a medium of lower index produces no phase change. Light passing into a new medium, refraction, never changes its phase. At the front of a soap film, light in air reflects from water, a higher index, so it is flipped. At the back, light in water reflects from air, a lower index, so it is not.

The second difference is distance. The back reflection travels through the film and back again, an extra path of 2t for a film of thickness t at normal incidence. Inside the film the light's wavelength is shorter, because it travels more slowly while its frequency stays the same:

λn = λ/n

and it is this wavelength in a medium, not the wavelength in air, that the extra path must be counted in. Red light of 650 nm has a wavelength of 650/1.333 = 488 nm in water.

Now the black film is explained. Where the film is far thinner than a wavelength, the extra path 2t is almost nothing, but the front reflection has been flipped by half a cycle and the back one has not. The two reflected waves are half a cycle out of step and cancel. Since that is true for every wavelength, all colours cancel together and the film looks black. OpenStax gives exactly this account of why a soap film darkens where it is thinnest, just before it breaks.

Bottom line: a very thin film is not too thin to reflect; it reflects twice, and the half-cycle flip at the front makes the two reflections cancel.

Counting phase changes and path together

For any thin film at normal incidence the recipe is the same. Count how many of the two reflections are flipped. If exactly one is, the flip already puts them half a cycle apart. Then they cancel when the extra path is a whole number of film wavelengths, and reinforce when it is a whole number plus a half. If neither or both are flipped, the flips do not separate them, so the path alone decides.

Here m = 0, 1, 2, .... OpenStax's soap bubble, water in air lit by 650 nm light, reflects strongly at thicknesses of about 122 nm, 366 nm and 610 nm, and not at all at 0, 244 nm and 488 nm, exactly as the first row predicts.

Worked problem: what colour is a 300 nm film?

Givens. A soap film 300 nm thick, with n = 1.333, surrounded by air, viewed by reflected white light at normal incidence.

The set-up in words. A flat film held upright in a wire loop, a lamp behind your shoulder, and you looking straight at the film, so the light strikes it and returns along the normal.

Principle. One phase flip, at the front surface, so bright reflection needs 2t = (m + ½)λ/n; solve for the wavelength and keep those between about 400 nm and 750 nm.

1. Rearrange: λ = 2nt/(m + ½) = 2(1.333)(300 nm)/(m + ½) = (800 nm)/(m + ½).
2. m = 0: 1600 nm, infrared. m = 1: 800 nm/1.5 = 533 nm, green. m = 2: 320 nm, ultraviolet.
3. The only visible wavelength reinforced is 533 nm, so the film looks green. Dark reflection, 2t = mλ/n, falls at 800 nm and 400 nm, the very edges of the visible range.

Sanity check. Check the bright case directly. In the film, 533 nm light has a wavelength of 400 nm, and the extra path of 600 nm is 1.5 of those. That is a whole number plus a half, which with one flip means reinforcement. A few hundred nanometres is the right thickness for interference colours: a film thousands of times thicker would reinforce so many closely spaced wavelengths that they would blend to white.

The variant that flips the result. Lay the same 300 nm layer of water on a glass plate, n = 1.52, instead of leaving it in air. Now the back reflection also comes off a higher index, water to glass, so both reflections are flipped and the flips cancel. The bright condition becomes 2t = mλ/n, giving 800 nm and 400 nm, while 533 nm now satisfies the dark condition. The green that the free film reflected strongly is the colour this layer removes. And where such a layer thins toward zero it does not go black at all, because with both reflections flipped the two waves arrive in step.

Colours in bubbles and oil films

A real soap film or bubble is not one uniform thickness. Liquid drains down a vertical film, so it is thinnest at the top and thicker toward the bottom, and on a bubble the thickness swirls as the liquid moves. Each thickness reinforces its own set of wavelengths and suppresses others. So the bands of colour follow the contours of equal thickness and change as the film drains, ending in the black band where it is thinnest. A thin film of oil on a wet road works the same way, its colours mapping its thickness. The colours also shift as you tilt your head, because the path through the film depends on the angle at which light crosses it. The exam limits calculations to light at normal incidence.

Coatings that cancel reflections

The same physics, run deliberately, removes unwanted reflections. Every surface of a glass lens reflects a few percent of the light that reaches it. In a camera with many lenses, that adds up to stray light and lost brightness. OpenStax's example coats glass (n = 1.52) with magnesium fluoride (n = 1.38). Because the coating's index lies between those of air and glass, both reflections come off a higher index and both are flipped, so only the path matters. For cancellation at 550 nm, in the middle of the visible range, the thinnest coating makes the round trip half a wavelength in the coating: 2t = λn/2, so t = λ/(4n) = 550 nm/(4 × 1.38) = 99.6 nm. That is a quarter of the wavelength in the coating. It is why the College Board describes the simplest anti-reflection coating as a quarter-wave layer, with an index greater than air's and less than the surface's. It cancels perfectly only at one wavelength and one angle, and it reduces the reflection of nearby colours well.

Where the reasoning broke

The "too thin to reflect" explanation failed at its first step. It treated reflection as something a layer does in proportion to how much of it there is, when reflection happens at each boundary regardless of thickness. Having assumed the reflections away, it never asked whether two reflections could cancel. The correct account keeps both reflections, notices that one is flipped by half a cycle and the other is not, and lets the extra path decide everything else. It also predicts what the wrong account cannot. The black appears only where one reflection is flipped, so a thin water layer on glass, flipped at both surfaces, brightens instead of going black as it thins.

Common misconceptions

• "A very thin film is too thin to reflect light." Each surface reflects. A very thin soap film looks black because its two reflections cancel.
• "Reflection always flips the phase." Only reflection from a higher-index medium does. The back of a soap film, water to air, reflects without a flip.
• "Light changes phase when it refracts into the film." Refraction never shifts the phase; only the reflections and the extra path matter.
• "The film's thickness should be compared with the wavelength in air." Inside the film the wavelength is λ/n, so the path is counted in film wavelengths.
• "Soap bubbles are coloured by dyes in the soap." The colours are interference, set by the film's thickness, which is why they move as the bubble drains.

What to remember

A soap film turns black at the top just before it bursts not because it is too thin to reflect but because its two reflections cancel. The front reflection, from a higher index, is flipped by half a cycle; the back one, from a lower index, is not. When the film is far thinner than a wavelength the extra path is negligible, so every colour cancels. In general, count the flips, then compare the extra path 2t with the wavelength in the film, λ_n = λ/n. With one flip, the film is bright when 2t = (m + ½)λ_n; with two or none, it is bright when 2t = mλ_n. A 300 nm soap film in air reflects green, 533 nm, while the same layer on glass removes that green and does not darken as it thins. Varying thickness gives bubbles and oil films their moving bands of colour.

A quarter-wave coating with an index between air and glass, 99.6 nm of magnesium fluoride for 550 nm, cancels a lens's reflection.

Sources

1. OpenStax. (2022). 27.7 Thin film interference (phase change on reflection; dark thin soap films; the soap bubble and the anti-reflection coating). In College Physics 2e. openstax.org
2. Nave, R. (n.d.). Thin film interference. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
3. OpenStax. (2022). 25.3 The law of refraction (indices of water, crown glass and magnesium fluoride). In College Physics 2e. openstax.org
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 14.9, Thin-film interference (quantitative analysis at normal incidence only). apcentral.collegeboard.org

A formula that was a lucky guess

On 19 October 1900, Max Planck showed a meeting of the German Physical Society in Berlin a formula for how the light given off by a hot object is spread across wavelengths. It fitted the best measurements of the day, and it was, as APS News recounts, hailed as indisputably correct. Planck himself called it a lucky guess. He had found it by combining two expressions in the simplest way that would match the data, and he did not know why it worked. By 14 December 1900 he had a derivation, and it came at a price he did not like. He had to assume that the vibrating charges inside the hot object could take in and give out energy not continuously, but only in discrete amounts, which he called quanta.

That assumption was the beginning of quantum physics, and this module follows it from the glow of hot objects to the inside of the nucleus. This first lesson is about the glow itself. It covers what spectrum a hot object gives out and the two laws that describe how that spectrum changes with temperature. Then it asks why the physics of 1900 could not explain it, and meets the particle of light, the photon, that Planck's quanta eventually became.

Everything glows

Matter spontaneously converts some of its internal thermal energy into electromagnetic radiation. The coil of an electric hob glows dull red and then orange as it heats; the Sun glows white; you glow too, although at body temperature the radiation is infrared, which your eyes cannot see but a thermal camera can. To study this radiation cleanly, physicists use an idealised model, the blackbody: an object that absorbs all the radiation falling on it, reflecting none. A blackbody in equilibrium at a constant temperature must also emit, and the radiation it emits has a continuous spectrum that depends only on its temperature, not on what it is made of.

The spectrum is shown as a graph of intensity per unit wavelength (vertical) against wavelength (horizontal). Picture it: near zero at very short wavelengths, rising steeply to a peak, then falling away more gradually into a long tail at long wavelengths. Raise the temperature and two things happen at once. The whole curve rises, at every wavelength, and its peak moves to shorter wavelengths. A hob ring at about 1000 K peaks in the infrared and gives out only a little red light from the short-wavelength edge of its curve; the Sun's peak lies in the middle of the visible range. OpenStax notes that the shape of this curve cannot be described with classical physics. Applying the classical theory of radiation to it predicted intensities that grow without limit toward short wavelengths, a failure that became known as the ultraviolet catastrophe. Planck's quantum assumption removed it.

Wien's law: the peak moves with temperature

The wavelength at which a blackbody emits the most radiation per unit wavelength, λ_max, is inversely proportional to its absolute temperature. This is Wien's law:

λmax = b/T

Wien's constant b is 2.897 771 955 × 10^-3 m·K in NIST's CODATA list, a value fixed by the exactly defined constants of the SI; the exam's table rounds it to 2.90 × 10^-3 m·K.

Givens. NASA's Sun fact sheet gives the Sun's effective surface temperature as 5772 K. Your skin is at about 310 K.

The set-up in words. Two glowing objects, treated as blackbodies: the Sun's surface, and a person in a dark room.

Principle. Wien's law, with temperatures in kelvin.

1. The Sun: λmax = (2.898 × 10-3 m·K)/(5772 K) = 5.02 × 10-7 m = 502 nm, in the green-blue part of the visible spectrum.
2. A person: λmax = (2.898 × 10-3 m·K)/(310 K) = 9.35 × 10-6 m, about 9 µm, deep in the infrared, which is the band thermal cameras are built to see.

Sanity check. The person is about 19 times cooler, so the peak should be about 19 times longer: 502 nm × 18.6 = 9.3 µm. The Sun's peak falls inside the range our eyes respond to, which is not a coincidence for eyes that evolved in sunlight.

The variant that flips the result. Take two stars. A cool star at 3000 K peaks at 966 nm, in the infrared, and looks red, because the visible part of its curve is strongest at the red end. A hot star at 10 000 K peaks at 290 nm, in the ultraviolet, and looks blue-white, because the visible part of its curve now rises toward the violet. Moving the temperature across the Sun's flips the peak out of the visible on one side or the other, and it is why red stars are the cool ones, the reverse of the everyday habit of calling red "hot" and blue "cold".

The Stefan-Boltzmann law: the total power

The total power radiated by a blackbody, over all wavelengths, is proportional to its surface area and to the fourth power of its absolute temperature:

P = AσT4

with the Stefan-Boltzmann constant σ = 5.670 374 419 × 10^-8 W/(m²·K⁴) from NIST, rounded to 5.67 × 10^-8 on the exam's table. The fourth power makes temperature dominant: double it and the power rises sixteen-fold. A hob ring at 1000 K radiates (1000/300)⁴ = 123 times as much per square metre as the same ring at room temperature.

The Sun makes a good test. NASA gives its mean radius as 695 700 km, so its surface area is 4πr² = 4π(6.957 × 10⁸ m)² = 6.08 × 10¹⁸ m², and

P = (6.08 × 1018 m2)(5.67 × 10-8 W/(m2·K4))(5772 K)4 = 3.83 × 1026 W

which is the luminosity on NASA's fact sheet, 382.8 × 10²⁴ J/s. Treated as a blackbody, the Sun's surface temperature and size account for all of its power.

Why this matters: the colour of a glowing object reports its temperature through Wien's law, and its brightness reports its temperature and size through the Stefan-Boltzmann law. Astronomers read stars this way.

Planck's step: energy comes in packets

Planck explained the blackbody curve by treating the charges in the walls of a hot object as oscillators whose energy could change only in steps of hf, where f is the frequency of the radiation and h is the constant now named after him. NIST lists the Planck constant as exactly 6.626 070 15 × 10^-34 J·s, one of the constants used since 2019 to define the SI. The steps are tiny for low frequencies and large for high ones. At a given temperature, oscillators have plenty of energy to emit infrared quanta, but a single ultraviolet quantum costs more energy than they usually have, so high-frequency emission is choked off. That is why the real curve turns down at short wavelengths instead of rising without limit, and why the catastrophe never happens.

Planck thought the quanta were a property of the emitting matter. In the next lesson's experiment, the photoelectric effect, it became clear that light itself comes in packets. Quantum theory was developed because three sets of observations defied classical physics: the blackbody spectrum, the photoelectric effect, and the sharp lines in the spectra of atoms, all of which this module explains. Quantum theory is needed to describe matter at atomic and subatomic scales, and in it fundamental particles show both wave-like and particle-like behaviour.

The photon

A photon is a particle of light: massless, electrically neutral, and carrying an energy proportional to its frequency,

E = hf = hc/λ

Photon energies are usually quoted in electron volts, and a convenient combination is hc = 1240 eV·nm. A red photon of 650 nm carries 1240/650 = 1.91 eV; a green photon of 500 nm, 2.48 eV (3.97 × 10^-19 J); a violet one of 400 nm, 3.10 eV. Photons travel in straight lines unless they interact with matter, at c in empty space and at c/n in a medium of index n. Light can be described as waves, as Module 6 did, or as a stream of photons, and which description you need depends on the question being asked.

The particle picture puts numbers on a beam. A 1.0 mW green laser pointer at 532 nm, each of whose photons carries hc/λ = 3.73 × 10^-19 J, emits (1.0 × 10^-3 J/s)/(3.73 × 10^-19 J) = 2.7 × 10¹⁵ photons every second. Ordinary light is so many photons that its graininess is invisible; that is why the wave model works so well for it.

Common misconceptions

• "Only hot objects give off radiation." Every object above absolute zero radiates. At room and body temperature the radiation is infrared.
• "Red stars are hotter than blue stars." The reverse: by Wien's law a hotter body peaks at a shorter wavelength.
• "Doubling the temperature doubles the power radiated." P goes as T⁴, so it rises sixteen-fold.
• "A brighter beam has more energetic photons." Each photon's energy is set by its frequency. A brighter beam of the same colour simply has more photons per second.
• "Planck's quanta made the blackbody curve rise faster at short wavelengths." They did the opposite: costly high-frequency quanta suppress short-wavelength emission and bend the curve down.

The short version

Planck's 1900 formula for the glow of hot objects was hailed as correct before he could explain it. The explanation he found by December assumed that energy is exchanged in discrete quanta of size hf, with h = 6.626 070 15 × 10^-34 J·s exactly. Every object radiates; an ideal blackbody gives a continuous spectrum fixed by its temperature alone, which classical physics could not reproduce. Wien's law, λ_max = b/T, puts the Sun's peak at 502 nm for 5772 K and yours near 9 µm. It makes 3000 K stars red and 10 000 K stars blue-white.

The Stefan-Boltzmann law, P = AσT⁴, gives the Sun's whole output of 3.83 × 10²⁶ W from its size and temperature, and makes power climb sixteen-fold when temperature doubles. Light comes as photons, massless and neutral, with E = hf = hc/λ: 1.91 eV for 650 nm red and 3.10 eV for 400 nm violet. Photons travel at c in vacuum and c/n in matter, and a 1.0 mW green pointer emits 2.7 × 10¹⁵ of them every second.

Sources

1. APS News. (2002, October). This month in physics history: Planck's quantum hypothesis (19 October and 14 December 1900). American Physical Society. aps.org
2. OpenStax. (2022). 29.1 Quantization of energy (blackbody radiation; the ultraviolet catastrophe; Planck); 29.3 Photon energies and the electromagnetic spectrum. In College Physics 2e. openstax.org
3. National Institute of Standards and Technology. (2024). CODATA 2022 values: Planck constant (exact); Wien wavelength displacement law constant; Stefan-Boltzmann constant. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
4. NASA Space Science Data Coordinated Archive. (n.d.). Sun fact sheet (radius, luminosity, effective temperature). nssdc.gsfc.nasa.gov
5. National Institute of Standards and Technology. (n.d.). SI redefinition (seven defining constants, among them the Planck constant, in effect from 20 May 2019). nist.gov
6. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 15.1 and 15.4. apcentral.collegeboard.org
7. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Blackbody spectrum (optional: set a temperature and watch the curve and its peak). phet.colorado.edu

A careful experiment by a sceptic

In March 1916, Robert Millikan published in Physical Review the most precise test yet of an equation he did not believe. Einstein had proposed it in 1905 to explain how light knocks electrons out of metals, by treating light as a stream of energy packets. Millikan, as the historian of science Gerald Holton recounts, called that idea "the bold, not to say the reckless, hypothesis".

He built an evacuated glass apparatus, a "machine shop in vacuo". Inside it, a knife worked by an electromagnet from outside scraped each metal surface clean just before it was lit. Then he measured the energies of the escaping electrons for one colour of light after another. The equation fitted. From it Millikan obtained Planck's constant, 6.57 × 10^-34 J·s, with a precision of about half a percent: within 1 percent of the value now fixed exactly by the SI.

That experiment is the subject of this lesson, worked through as a procedure. First come the observations that the wave model of light could not explain, then Einstein's equation. Then one metal and one colour are worked end to end, and finally comes the straight-line graph from which Millikan read h.

What the experiment does

The photoelectric effect is the emission of electrons from a material when electromagnetic radiation falls on it. The College Board describes the standard apparatus: two metal plates in a vacuum chamber, connected to a variable source of potential difference. One plate is lit with monochromatic light, light of a single frequency. If the light frees electrons from that plate, some cross to the other plate and a current, the photocurrent, flows in the circuit.

Now make the collecting plate negative, so that it repels the arriving electrons. Increase that reverse potential difference until even the fastest electrons are turned back and the current stops. That is the stopping potential, ΔV_s. An electron that only just fails to arrive has lost all its kinetic energy climbing the potential difference, exactly as the electron in Lesson 9 did. So the maximum kinetic energy of the emitted electrons is

Kmax = eΔVs

Three observations the wave model could not explain

If light were only a wave, its energy would be spread smoothly over the metal's surface. Brighter light would deliver energy faster, and any colour would eventually free electrons if it were bright enough. OpenStax's University Physics lists what experiments actually show.

• A threshold frequency. Below a certain frequency, characteristic of the metal, no electrons are emitted at all, however bright the light. Above it, they are emitted however dim the light.
• Intensity changes the number, not the energy. Brighter light of the same frequency gives more electrons, a bigger photocurrent, but the maximum kinetic energy, and so the stopping potential, stays the same. Raising the frequency raises the maximum kinetic energy.
• No delay. Electrons are emitted almost instantaneously, even at very low intensities, with none of the long wait a spread-out wave would need to deliver enough energy to one electron.

Einstein's photoelectric equation

All three follow at once if light arrives as photons of energy hf, each absorbed whole by a single electron. The electron must spend some energy to escape the metal; the least it can spend is the work function φ, a property of the metal. Whatever is left over is kinetic energy, so the most energetic electrons have

Kmax = hf - φ

If hf is less than φ, no single photon can free an electron, so there is a threshold frequency f₀ = φ/h, or equivalently a cutoff wavelength λ₀ = hc/φ. Brighter light is more photons, so more electrons but no more energy each. And since each photon acts on one electron the moment it arrives, there is no delay. The exam provides work functions when it needs them; for reference, OpenStax's University Physics lists typical values:

The procedure, part one: one metal, one colour

Givens. Ultraviolet light of wavelength 254 nm falls on a clean zinc plate, work function 4.31 eV. Use hc = 1240 eV·nm and m_e = 9.11 × 10^-31 kg.

The set-up in words. The zinc plate and a collecting plate face each other in a vacuum tube, wired through a meter to an adjustable supply that can make the collector negative.

Principle. Photon energy E = hc/λ, Einstein's equation, and K_max = eΔV_s.

1. Photon energy: E = hc/λ = (1240 eV·nm)/(254 nm) = 4.88 eV.
2. Maximum kinetic energy: Kmax = 4.88 eV - 4.31 eV = 0.57 eV.
3. Stopping potential: 0.57 eV per electron of charge e means ΔVs = 0.57 V.
4. Fastest electrons: v = √(2K/m) = √[2(0.57)(1.60 × 10-19 J)/(9.11 × 10-31 kg)] = 4.5 × 105 m/s.

Sanity check. The photon energy exceeds the work function, so emission is possible, and the leftover energy is a fraction of a volt, the right size for stopping potentials in such tubes. Electron volts convert straight into volts for a single electron, which is why the stopping potential and K_max share a number.

The variant that flips the result. Shine a red laser of 650 nm on the same zinc instead. Each photon now carries 1240/650 = 1.91 eV, less than half the work function. No electrons come out, and none will however bright the laser is made. Brightness adds photons, not energy per photon, and an electron cannot save up energy from several photons. Ultraviolet light of modest intensity frees electrons; red light of any intensity frees none. On the wave picture that result is incomprehensible; on the photon picture it is the first thing you would predict.

So what?: the energy of each photoelectron is set by the light's frequency and the metal's work function; the number of photoelectrons is set by the light's intensity.

The procedure, part two: the graph Millikan drew

Divide Einstein's equation by e and it becomes a straight line: ΔVs = (h/e)f - φ/e. Plot stopping potential (vertical) against frequency (horizontal) for one metal, and the slope is h/e, the same for every metal. The line crosses the frequency axis at the threshold frequency. Here are illustrative readings for sodium:

The five points with current lie on a straight line. Its slope is (1.06 V - 0.23 V)/(8.5 - 6.5) × 10¹⁴ Hz = 4.15 × 10^-15 V/Hz. So h = e × slope = (1.60 × 10^-19 C)(4.15 × 10^-15 V·s) = 6.6 × 10^-34 J·s. Extended back, the line reaches zero at about 5.95 × 10¹⁴ Hz, the threshold frequency, which is why the 5.5 × 10¹⁴ Hz light gave no current. The work function follows: φ = hf₀ = (4.14 × 10^-15 eV·s)(5.95 × 10¹⁴ Hz) = 2.46 eV, matching the table. Repeat with zinc and you get a parallel line, with the same slope and a threshold of 1.04 × 10¹⁵ Hz, in the ultraviolet. A slope that is the same for every metal, and equal to a constant first met in blackbody radiation, is what made the photon hard to dismiss.

Running it backwards: h from coloured LEDs

A light-emitting diode does roughly the reverse. An electron driven across the junction inside the LED gives up its energy as a photon. An LED begins to emit noticeably once the potential difference across it gives each electron about as much energy as one photon carries. So e times the activation voltage is roughly hc/λ. Maria André and Paulo André, writing in Science in School in 2014, give these typical values:

A graph of activation voltage against 1/λ is close to a straight line of slope hc/e. Their four points give a slope of 1.25 × 10^-6 V·m, and so h = (1.60 × 10^-19 C)(1.25 × 10^-6 V·m)/(3.00 × 10⁸ m/s) = 6.7 × 10^-34 J·s. It is a cruder method than Millikan's. An LED emits a narrow band of wavelengths rather than one, and the turn-on point has to be judged, but the method needs only a battery, a potentiometer and two meters.

Common misconceptions

• "Brighter light gives faster electrons." Brighter light gives more electrons. Their maximum energy depends only on the frequency and the metal.
• "Any light will free electrons if it is bright enough." Below the threshold frequency no electrons are emitted at any intensity: red light never frees electrons from zinc.
• "Every emitted electron has the energy hf - φ." That is the maximum, for electrons that spend only the minimum energy escaping. Most have less.
• "A higher stopping potential means a brighter light." The stopping potential measures K_max, which brightness does not change.
• "The photoelectric effect proves light is only particles." It shows light delivers energy in photons. Interference shows it also behaves as a wave, and both descriptions are needed.

Where this leaves us

Millikan set out to test an equation he thought reckless and confirmed it, getting h = 6.57 × 10^-34 J·s within about half a percent. In the photoelectric effect, light frees electrons from a metal. The threshold frequency, the independence of electron energy from intensity, and the absence of any delay all defy the wave model. They follow from photons of energy hf, each absorbed whole by one electron: K_max = hf - φ, measured as eΔV_s by the stopping potential. Ultraviolet at 254 nm frees electrons from zinc (φ = 4.31 eV) with 0.57 eV to spare, while 650 nm red light, at 1.91 eV per photon, frees none however bright.

A graph of stopping potential against frequency is a straight line of slope h/e for every metal. It crosses the axis at the threshold frequency, 5.95 × 10¹⁴ Hz for sodium. The reverse process in LEDs gives a rough h of 6.7 × 10^-34 J·s from four coloured diodes. The next lesson shows that photons carry momentum as well as energy.

Sources

1. Holton, G. (1999). Centennial focus: Millikan's measurement of Planck's constant. Physical Review Focus, 3, 23. American Physical Society. physics.aps.org
2. Millikan, R. A. (1916). A direct photoelectric determination of Planck's "h". Physical Review, 7(3), 355. https://doi.org/10.1103/PhysRev.7.355. Abstract page: journals.aps.org
3. OpenStax. (2016). 6.2 Photoelectric effect (Table 6.1, work functions; the observations that defy classical theory). In University Physics Volume 3. openstax.org
4. André, M. R. A. S. F., and André, P. S. B. (2014). Classroom fundamentals: Measuring the Planck constant. Science in School, issue 28. scienceinschool.org
5. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 15.5 (work functions are provided on the exam). apcentral.collegeboard.org
6. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Photoelectric effect (optional: vary colour, intensity and voltage). phet.colorado.edu

X-rays that come out with a longer wavelength

In the early 1920s Arthur Compton and his collaborators aimed X-rays of a single wavelength at a block of graphite and measured the wavelengths of the X-rays scattered off it at different angles. Some came out unchanged. Others came out with a longer wavelength than they went in with, and the further they had been deflected, the longer the wavelength: an X-ray turned through 90° gained 2.4 picometres, one turned almost straight back nearly 5. Compton published his explanation in 1923, and it made the photon impossible to treat as a mere bookkeeping device. The puzzle to solve is why a scattered X-ray should change its wavelength at all.

Why a wave should not do that

On the wave model, an X-ray is an oscillating electric field. When it passes over an electron, the field shakes the electron back and forth at the wave's own frequency, and the shaking electron radiates a new wave, the scattered wave, at that same frequency. The scattered X-rays should have the same frequency, and so the same wavelength, as the incoming ones, in every direction. The longer wavelengths Compton measured, depending neatly on angle, do not fit this picture at all.

Photons carry momentum

They fit perfectly if X-rays are photons that collide with electrons like particles. A photon carries energy E = hf, and, though it has no mass, it also carries momentum:

p = h/λ

Now picture the collision. A photon strikes a nearly free electron at rest. Momentum is conserved in two dimensions, so if the photon is deflected, the electron must recoil to balance the sideways momentum. Energy is conserved too, so the kinetic energy the electron recoils with must come out of the photon. A photon with less energy has a lower frequency and a longer wavelength. Working the two conservation laws through, as Compton did, gives the shift in wavelength:

Δλ = λ' - λ = (h/(mec))(1 - cosθ)

where θ is the angle through which the photon is scattered. The combination h/(m_ec) is the electron's Compton wavelength, which NIST's CODATA 2022 list gives as 2.426 310 235 38 × 10^-12 m, about 2.43 pm. Notice what the formula says. The shift depends only on the angle, not on the wavelength of the incoming photon. It is zero for a photon that carries straight on, 2.43 pm at 90°, and largest, 4.85 pm, for a photon bounced straight back.

Compton scattering is evidence that light is a stream of discrete packets. Treating each photon as a particle and applying conservation of energy and conservation of momentum to its collision with one electron predicts exactly what is measured.

Worked problem: an X-ray scattered through 90°

Givens. An X-ray photon of wavelength 71.0 pm strikes a free electron at rest and is scattered through 90°. Use h = 6.63 × 10^-34 J·s, hc = 1240 eV·nm and h/(m_ec) = 2.43 pm.

The set-up in words. The photon arrives travelling along the +x axis and leaves travelling along +y. The electron, initially at rest at the origin, recoils somewhere below the x axis.

Principle. The Compton formula for the new wavelength, energy conservation for the electron's kinetic energy, and momentum conservation in two dimensions for its direction.

1. New wavelength: λ' = λ + (2.43 pm)(1 - cos 90°) = 71.0 pm + 2.43 pm = 73.4 pm.
2. Photon energies: before, E = (1240 eV·nm)/(0.0710 nm) = 17.46 keV; after, E' = (1240 eV·nm)/(0.07343 nm) = 16.89 keV.
3. The electron's kinetic energy is the difference: K = 17.46 keV - 16.89 keV = 0.58 keV, about 577 eV.
4. Momentum along x: the photon brought h/λ = (6.63 × 10^-34)/(7.10 × 10^-11 m) = 9.33 × 10^-24 kg·m/s and leaves none in that direction, so the electron carries 9.33 × 10^-24 kg·m/s along +x.
5. Momentum along y: the scattered photon carries h/λ' = 9.02 × 10^-24 kg·m/s along +y, and the total was zero, so the electron carries 9.02 × 10^-24 kg·m/s along -y.
6. The electron's momentum: p = √(9.332 + 9.022) × 10-24 = 1.30 × 10-23 kg·m/s, at tan-1(9.02/9.33) = 44° below the x axis.

Sanity check. The two conservation laws must agree. From the momentum, the electron's kinetic energy is p²/(2m_e) = (1.30 × 10^-23)²/[2(9.11 × 10^-31)] J = 9.25 × 10^-17 J = 577 eV, the same as the energy difference. The photon lost about 3 percent of its energy, and its wavelength grew by the same 3 percent.

The variant that flips the result. Repeat with visible light of 500 nm. The shift at 90° is still 2.43 pm, because it does not depend on the wavelength, but now it is 2.43 pm out of 500 000 pm, a change of five parts in a million, far too small to notice. That is why the effect was found with X-rays: the Compton shift is a fixed size, so it is only visible when the wavelength itself is comparably small. And at θ = 0 the shift vanishes for every wavelength: a photon that is not deflected has given the electron nothing.

Key idea: photons carry momentum p = h/λ. In a collision with an electron they obey the same conservation laws as billiard balls, losing energy and so gaining wavelength by an amount set by the scattering angle.

If waves can be particles, particles can be waves

Louis de Broglie turned the argument round. If light, a wave, carries momentum p = h/λ, perhaps every particle with momentum p has a wavelength,

λ = h/p

now called its de Broglie wavelength. OpenStax works the decisive case. An electron accelerated through 54 V has 54 eV of kinetic energy and moves at about 4.36 × 10⁶ m/s, so its momentum is m_ev and its wavelength is 0.167 nm, comparable to the spacing of atoms in a crystal. Clinton Davisson and Lester Germer, scattering electrons from crystals, found exactly the diffraction patterns that waves of that wavelength would give; G. P. Thomson found the same independently. Electrons sent through two slits build up an interference pattern, just as light does in Young's experiment.

Why don't you diffract through a doorway? Compare a baseball of mass 0.145 kg thrown at 40 m/s. Its de Broglie wavelength is (6.63 × 10^-34 J·s)/[(0.145 kg)(40 m/s)] = 1.1 × 10^-34 m, unimaginably smaller than any opening or obstacle it could meet, so its wave nature never shows. Quantum theory is needed to describe a system when the de Broglie wavelength of its particles is comparable to the system's size. An electron in an atom, about 10^-10 m across, qualifies; a baseball in a stadium does not. The wavelength also grows as momentum falls, which is why slow electrons show their wave nature most plainly. A proton at the electron's 4.36 × 10⁶ m/s, 1836 times as massive, has a wavelength 1836 times shorter.

One object, two behaviours

This is wave-particle duality, and it is easy to misread. Photons and electrons are not sometimes waves and sometimes particles in the sense of switching. They are quantum objects that show wave-like behaviour, such as interference and diffraction, in experiments that test for it, and particle-like behaviour, such as definite energies and momenta exchanged in collisions, in experiments that test for that. Photons travel in straight lines unless they interact with matter, and when they interact, they do so as whole packets. One more consequence, which the next lesson uses: when a particle is confined, as an electron is in an atom, only certain standing waves fit, so a bound system can have only certain energies and momenta. In quantum theory the energies of bound systems are quantized.

Common misconceptions

• "A photon has no mass, so it cannot have momentum." It carries momentum p = h/λ, which is what the recoiling electron in Compton scattering receives.
• "The Compton shift is bigger for longer wavelengths." The shift depends only on the scattering angle, at most 4.85 pm; it is noticeable only for short wavelengths such as X-rays.
• "A scattered photon slows down." It still travels at c. It loses energy by lowering its frequency, which lengthens its wavelength.
• "Heavier particles have longer de Broglie wavelengths." λ = h/p, so more momentum means a shorter wavelength. That is why matter waves are seen for electrons, not baseballs.
• "Matter waves mean the electron is smeared out like a cloud of mass." The wave describes where the electron is likely to be found; each detection still finds a whole electron at one place.

Putting it together

Compton's X-rays came off graphite with longer wavelengths the further they were scattered, which a wave model cannot explain and a photon model predicts. A photon of momentum h/λ collides with an electron, conserves energy and momentum in two dimensions, and loses energy. So its wavelength grows by Δλ = (h/(m_ec))(1 - cosθ), with h/(m_ec) = 2.43 pm. A 71.0 pm X-ray scattered through 90° becomes 73.4 pm and sends the electron off at 44° below its original direction with 577 eV, the two conservation laws agreeing. For 500 nm light the same 2.43 pm shift is a change of five parts in a million.

Turning the idea round, de Broglie gave every particle a wavelength λ = h/p. It is 0.167 nm for a 54 eV electron, the size of atomic spacings, which is why crystals diffract electrons. For a thrown baseball it is 1.1 × 10^-34 m, which is why baseballs never diffract. Quantum theory is needed when a particle's wavelength is comparable to the system it is in, and confining a wave quantizes the energies of bound systems.

Sources

1. OpenStax. (2016). 6.3 The Compton effect (Compton's 1923 explanation; the shift formula; the Compton wavelength). In University Physics Volume 3. openstax.org
2. OpenStax. (2022). 29.4 Photon momentum; 29.6 The wave nature of matter (de Broglie; Davisson and Germer; the 54 eV electron). In College Physics 2e. openstax.org
3. National Institute of Standards and Technology. (2024). CODATA 2022 value: Compton wavelength. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
4. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 15.1 and 15.6 (full quantitative treatment of momentum conservation in two dimensions). apcentral.collegeboard.org
5. PhET Interactive Simulations, University of Colorado Boulder. (n.d.). Davisson-Germer: Electron diffraction (optional: fire electrons at a crystal and vary their speed). phet.colorado.edu

Two kinds of lines from one gas

Seal a little hydrogen in a glass tube, excite it with about 5000 volts, as in the demonstration HyperPhysics describes, and the gas glows. Look at that glow through a diffraction grating and you do not see a rainbow. You see a few separate bright lines: a red one at 656 nm, a blue-green one at 486 nm, and two violet ones at 434 nm and 410 nm. NIST's tables of strong hydrogen lines give them as 656.28, 486.13, 434.05 and 410.17 nm, measured in air.

Now turn to sunlight, spread into its rainbow (never by looking at the Sun; the experiment below shows a safe way). In 1815 Joseph Fraunhofer found about 600 dark lines crossing the solar spectrum. Dark lines and bright lines turned out to be two faces of one thing. Each element produces its own characteristic pattern of lines, and the wavelengths of an element's lines are the same whether they appear bright or dark. In 1860 Gustav Kirchhoff identified sodium in the Sun from its spectrum, and helium was found in the Sun's spectrum before anyone identified it on Earth.

This lesson explains both patterns with one model of the atom, and it is built as three comparisons: the planetary atom against the Bohr atom, emission against absorption, and a photon against a colliding electron. It covers AP Physics 2 Topics 15.2 and 15.3, which set a useful boundary: the exam describes atoms only through their energy levels, and draws energy-level diagrams only for atoms with a single electron. Orbital shapes and probability functions are left for later courses.

What an atom is made of

Lesson 9 followed Rutherford's alpha particles as they bounced off gold, and his conclusion still stands. An atom is a small, positively charged nucleus, about 10^-15 m across, surrounded by negatively charged electrons that fill a region about 10^-10 m across, so the nucleus is roughly 100 000 times smaller than the atom. The nucleus is made of protons and neutrons.

Two numbers describe a nucleus. The atomic number Z is the number of protons, and it fixes the element: every carbon nucleus has 6 protons, and a nucleus with 7 is nitrogen. The mass number A is the total number of protons and neutrons, so the number of neutrons is N = A - Z. Nuclear notation writes A at the upper left and Z at the lower left of the element's symbol: ¹²₆C is carbon with 6 protons and 6 neutrons, and ¹⁴₆C is carbon with 6 protons and 8 neutrons. Nuclei with the same Z but different numbers of neutrons are isotopes of the same element. How an atom interacts with other atoms depends on the number and arrangement of its electrons, and a neutral atom has Z electrons whatever its neutron count.

An atom with as many electrons as protons is neutral. An ion is an atom with a nonzero net charge: a sodium atom that has lost one electron is the ion Na⁺, with charge +e, and a chlorine atom that has gained one is Cl^-, with charge -e. Last, the mass. A proton is 1836 times as massive as an electron (CODATA 2022 gives m_p/m_e = 1836.15), and a neutron is slightly heavier than a proton, so the nucleus carries almost all of an atom's mass. The electrons, light as they are, set the atom's size and decide how it behaves, and they are what this lesson is about.

Comparison one: the planetary atom and the Bohr atom

Rutherford pictured a solar system in miniature: light electrons orbiting a heavy nucleus, with the Coulomb attraction doing the job gravity does for the planets. The picture has a fatal flaw. An electron moving in a circle is accelerating, and accelerated charges radiate electromagnetic waves. A radiating electron loses energy and spirals inward; as OpenStax puts it, classical orbits would decay quickly and matter would collapse. And a spiralling electron, its orbit shrinking as it goes, would radiate a continuous range of frequencies rather than the sharp lines a hydrogen tube actually shows.

In 1913 Niels Bohr kept the Coulomb force and the circular orbits but added rules that classical physics could not justify. Only certain orbits are allowed: those in which the electron's angular momentum is a whole-number multiple of h/(2π),

mevr = nh/(2π), with n = 1, 2, 3, ...

An electron in an allowed orbit does not radiate. Light is emitted or absorbed only when the electron jumps from one allowed orbit to another, and the photon carries the difference in energy. The table sets the two models side by side.

The Bohr model is a historical model, and the CED says so: it is built on classical physics, and it is the step that led to describing hydrogen by discrete energy states. Modern quantum theory keeps those energy states but drops the orbits; in OpenStax's words there are no well-defined orbits, only clouds of probability. For the exam, as for this lesson, the energy levels are what matter.

Worked problem: the ground state from Coulomb's law and circular motion

Givens. In the Bohr model of hydrogen, an electron (mass m_e = 9.11 × 10^-31 kg, charge -e) moves in a circle of radius r = 5.29 × 10^-11 m around a proton of charge +e. That radius is the Bohr radius a_B, which CODATA 2022 gives as 5.291 772 105 44 × 10^-11 m. Use k = 8.99 × 10⁹ N·m²/C² and e = 1.60 × 10^-19 C. Find the electron's speed, its kinetic energy, the electric potential energy of the electron-proton system, and the total energy.

Diagram in words. A dot for the proton at the centre and a circle of radius r around it. The electron sits on the circle, with its velocity arrow along the circle and its single force arrow pointing straight at the proton.

Principle. The Coulomb attraction is the only force, so it alone supplies the centripetal force: F_net = m_ev²/r. The energies follow from K = (1/2)m_ev² and U_E = kq₁q₂/r, which is negative here because the two charges have opposite signs.

Algebra with units.

1. Force: F = ke²/r² = (8.99 × 10⁹ N·m²/C²)(1.60 × 10^-19 C)²/(5.29 × 10^-11 m)² = 8.22 × 10^-8 N.
2. Speed: m_ev²/r = F, so v² = Fr/m_e = (8.22 × 10^-8 N)(5.29 × 10^-11 m)/(9.11 × 10^-31 kg) = 4.78 × 10¹² m²/s², and v = 2.19 × 10⁶ m/s.
3. Kinetic energy: K = (1/2)m_ev² = (1/2)Fr = (1/2)(4.35 × 10^-18 J) = 2.18 × 10^-18 J. Dividing by 1.60 × 10^-19 J per eV gives 13.6 eV.
4. Potential energy: U_E = -ke²/r = -Fr = -4.35 × 10^-18 J = -27.2 eV.
5. Total energy: E = K + U_E = 13.6 eV + (-27.2 eV) = -13.6 eV.

Sanity check. The total is negative, as it must be for a bound system: you would have to add 13.6 eV to pull the electron free, and 13.6 eV is the measured energy needed to ionize hydrogen. The speed is under 1 percent of the speed of light, so the ordinary formulas for kinetic energy are fine. Notice the pattern, which follows from m_ev² = ke²/r: the kinetic energy is exactly half the size of the potential energy, so the total energy is minus the kinetic energy.

The variant that flips the result. Put the electron in the n = 2 orbit instead, at r₂ = 4a_B = 2.12 × 10^-10 m. The force is 16 times weaker and Fr is 4 times smaller, so v² is 4 times smaller and the speed halves to 1.09 × 10⁶ m/s. Every energy shrinks by a factor of 4: K = 3.40 eV, U_E = -6.80 eV and E = -3.40 eV. The slower electron has the higher total energy, and only 3.40 eV will free it. Intuition says the fast electron is the one that escapes easily; the energy bookkeeping says the opposite. The ground state, where the electron moves fastest, is the hardest state to ionize, which is the CED's statement that an atom in its ground state needs the greatest energy to remove its electron.

Why only some orbits fit: the standing-wave picture

Bohr's rule looked arbitrary until 1924, when Louis de Broglie proposed that every particle has a wavelength λ = h/p (Lesson 28). Picture the electron's wave running round its orbit. It can only form a steady standing wave if a whole number of wavelengths fits exactly round the circumference, the same kind of condition that picks out the standing waves on a guitar string, but bent into a closed loop:

2πrn = nλ

Substituting λ = h/(m_ev) gives 2πr_nm_ev = nh, which is Bohr's rule m_evr = nh/(2π). Check it with the numbers just found. In the ground state, λ = h/(m_ev) = (6.63 × 10^-34 J·s)/[(9.11 × 10^-31 kg)(2.19 × 10⁶ m/s)] = 3.32 × 10^-10 m, and the circumference is 2π(5.29 × 10^-11 m) = 3.32 × 10^-10 m: exactly one wavelength fits. In the n = 2 orbit the speed halves, so the wavelength doubles to 6.65 × 10^-10 m, while the circumference grows four times to 1.33 × 10^-9 m: exactly two wavelengths fit. OpenStax's University Physics works the same check and finds λ = 2πa₀ ≈ 0.332 nm for the ground state.

This is the CED's standing-wave model: the circumference of the electron's orbit must be a whole number of de Broglie wavelengths, and that is why only certain energies are allowed. It is the clearest example of the idea Lesson 28 ended on: a confined wave can only take certain shapes, so a bound system can only have certain energies.

Reading an energy-level diagram

Carry the force-and-energy calculation through for every allowed orbit and the results fall into a pattern. For hydrogen,

En = -13.6 eV/n2 and rn = n2aB, with n = 1, 2, 3, ...

An energy-level diagram draws each allowed energy as a horizontal line. Height stands for energy and nothing else; the horizontal direction means nothing. The zero of energy is chosen where the electron is free and at rest, infinitely far from the proton, so every bound state has a negative energy. The lowest level, n = 1, is the ground state, and the others are excited states. The levels crowd together as n grows and converge on 0 eV, the ionization limit.

The energy needed to remove the electron from a given level is its ionization energy; the CED calls it the electron's binding energy. It is largest from the ground state. The atom is a system of a nucleus and an electron, and a change of level changes the energy of that system, so the difference has to come from somewhere or go somewhere. Usually it is a photon:

Ephoton = hf = hc/λ = |Eupper - Elower|

A drop from a higher level to a lower one emits a photon carrying exactly the difference. A photon carrying exactly the difference can be absorbed, lifting the atom from the lower level to the higher one. With hc = 1240 eV·nm, the wavelength in nanometres is 1240 divided by the energy gap in electron volts. Because each gap has a fixed size, each transition gives a single frequency and so a single wavelength: a line.

Remember: a photon is absorbed only if its energy matches the gap between two levels exactly; photons that match no gap stream past the atoms untouched. The one exception is ionization. Any photon carrying 13.6 eV or more can ionize a ground-state hydrogen atom, because a freed electron can leave with any amount of kinetic energy, so the surplus becomes the electron's kinetic energy, just as in the photoelectric effect.

Worked problem: hydrogen's red line, and a jump you cannot see

Givens. A hydrogen atom drops from n = 3 (E₃ = -1.51 eV) to n = 2 (E₂ = -3.40 eV). Use hc = 1240 eV·nm. Find the photon's energy and wavelength, and decide whether you could see it.

Diagram in words. The energy-level ladder from the table, with a downward arrow from the n = 3 rung to the n = 2 rung and a wavy arrow leaving to the side for the photon.

Principle. Conservation of energy: the atom's energy falls by E₃ - E₂, and the photon carries that energy away.

Algebra with units. E_photon = E₃ - E₂ = (-1.51 eV) - (-3.40 eV) = 1.89 eV. Then λ = hc/E_photon = (1240 eV·nm)/(1.89 eV) = 656 nm.

Sanity check. The energy is positive and a couple of electron volts, the size of a visible photon. 656 nm is red light, and NIST lists hydrogen's strongest visible line at 656.28 nm. A model built from Coulomb's law and one quantum rule has predicted a measured wavelength to within a fraction of a percent.

The variant that flips the result. Start from the same level, n = 3, but let the atom drop all the way to n = 1. Now E_photon = (-1.51 eV) - (-13.6 eV) = 12.09 eV and λ = (1240 eV·nm)/(12.09 eV) = 103 nm, deep in the ultraviolet. Same starting level, different landing level, and the line vanishes from sight. That is why the visible hydrogen lines all end on n = 2: drops to n = 1 release too much energy to be visible, and drops to n = 3 too little.

Lines are grouped into series by the level they end on. The Lyman series ends on n = 1 and lies entirely in the ultraviolet. The Balmer series ends on n = 2; part of it is visible, and the rest crowds towards 365 nm in the near ultraviolet. The Paschen series ends on n = 3 and, like all the series beyond it, lies entirely in the infrared: its first line, 4 to 3, has a gap of 0.66 eV and a wavelength of about 1880 nm. The table compares the Bohr predictions with measurement.

An atom excited to n = 4 can come down by several routes: straight to 1, or by way of 3 or 2. So it can produce six different lines: 4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1 and 2 to 1. In general, n levels allow n(n - 1)/2 different lines, one for each pair of levels. A gas holds a vast number of atoms, each making its own jumps, so a glowing tube shows all its lines at once.

Comparison two: emission and absorption

Emission and absorption are the same transitions run in opposite directions. In an emission spectrum, a hot, thin gas whose atoms are being excited, by an electric discharge or by heating, emits photons as the atoms fall back down, and you see bright lines on a dark background. In an absorption spectrum, a cooler, thin gas sits between you and a source of continuous light; its atoms take out exactly the photons that lift them from one level to a higher one, and you see a continuous rainbow crossed by dark lines. Hot, dense matter, such as a lamp filament or the body of a star, gives the continuous spectrum itself.

The absorbing atoms do not keep the energy for long: an atom typically stays excited for about a hundred-millionth of a second, then drops back and emits. The dark lines stay dark anyway, because the atoms re-emit in all directions and only a small fraction of that light continues along the original beam.

Worked problem: what a cool cloud of hydrogen takes out of white light

Givens. A cloud of cool hydrogen, with practically every atom in the ground state, sits in front of a source of continuous light covering 90 nm to 700 nm. Which wavelengths are missing from the light that comes through, and are any of them visible?

Diagram in words. On the left, the continuous source; in the middle, the cloud; on the right, a grating spectrometer. Beside it, the energy-level ladder with upward arrows that all start from n = 1, the only level the atoms occupy.

Principle. An atom absorbs a photon only if the photon's energy equals the gap from the atom's level to a higher level, or is at least the ionization energy.

Algebra with units. From n = 1 the gaps are 13.6 - 3.40 = 10.2 eV, 13.6 - 1.51 = 12.09 eV, 13.6 - 0.850 = 12.75 eV, and so on, closing in on 13.6 eV. The matching wavelengths are (1240 eV·nm)/(10.2 eV) = 122 nm, (1240 eV·nm)/(12.09 eV) = 103 nm and (1240 eV·nm)/(12.75 eV) = 97.3 nm, crowding towards (1240 eV·nm)/(13.6 eV) = 91.2 nm. Every photon shorter than 91.2 nm carries more than 13.6 eV and can ionize the atoms, so that whole region is absorbed too.

Sanity check. Every missing wavelength is shorter than 122 nm, in the ultraviolet, so the visible light comes through untouched: cool hydrogen shows no visible absorption lines at all. That fits the energies: the smallest jump up from the ground state needs 10.2 eV, more than five times the energy of a red 656 nm photon.

The variant that flips the result. Heat the cloud so that collisions keep a share of the atoms in n = 2. Those atoms can absorb photons of 1.89, 2.55, 2.86 and 3.02 eV, so dark lines appear at 656, 486, 434 and 410 nm, the visible Balmer lines. OpenStax's astronomy text makes the same point: absorption at 656 nm needs atoms whose electrons are already on the second level. Heat the gas much more and the result flips back. Once the atoms are ionized, a bare proton has no electron left to jump, and ionized hydrogen produces no absorption lines.

Comparison three: a photon and a colliding electron

Photons are not the only way to excite an atom. In a discharge tube, and in any hot gas, atoms are also excited by collisions. The rules differ in one important way. A photon is absorbed whole or not at all, so below the ionization energy it must match a gap exactly. A colliding electron can hand over just the energy of one gap and carry on with the rest.

James Franck and Gustav Hertz showed this in 1914. They accelerated electrons through mercury vapour and measured the current reaching a collector. As the accelerating voltage rose, the current rose, until at 4.9 V it dropped sharply: electrons with 4.9 eV were giving it up to mercury atoms in collisions and could no longer reach the collector. The drops came again at multiples of 4.9 V, as electrons with enough energy gave up 4.9 eV more than once on the way. The excited mercury atoms emit ultraviolet light at 254 nm, and hc/λ = (1240 eV·nm)/(254 nm) = 4.88 eV: the same gap, measured two different ways.

Apply this to hydrogen in its ground state. A 12.5 eV photon matches no gap (the nearest are 12.09 eV and 12.75 eV), so it passes straight through. An electron with 12.5 eV of kinetic energy can lift the atom to n = 2, keeping 2.3 eV, or to n = 3, keeping 0.41 eV, but it cannot reach n = 4, which needs 12.75 eV. The same 12.5 eV, two different outcomes.

Try it: a CD spectroscope at home, a discharge tube at school

You can see the difference between a continuous spectrum and a line spectrum with cardboard and an old CD, whose closely spaced tracks act as the reflection grating of Lesson 24. The idea is simple: a narrow slit, less than 1 mm wide and a few centimetres long, at one end of a cardboard tube or box, and the CD at the other end, tilted so that light from the slit reflects off its shiny side into a viewing hole. Swiss Science Center Technorama publishes a cutting guide for a tube version, with the CD at about 30 degrees to the end of the tube; NASA's STEREO mission publishes a cereal-box version that uses a diffraction grating instead.

1. Point the slit at a fluorescent tube or a compact fluorescent bulb that is already fitted and switched on. Look in at the CD and tilt the spectroscope until a spectrum appears beside the image of the slit.
2. Record the colours and the order of any bright lines, and sketch their spacing.
3. Compare an incandescent bulb if you can find one, and sunlight reflected off a sheet of white paper. Other lamps, such as LED bulbs and street lights, are worth a look too; record what you see.

The incandescent filament gives a continuous spectrum, one colour shading into the next. The fluorescent lamp gives a few sharp, bright coloured lines on a fainter continuous band: the lines come from the mercury vapour inside the tube, and the continuous band from the phosphors coating its inside wall. NIST's tables put mercury's strongest visible lines at about 405 nm (violet), 436 nm (blue), 546 nm (green) and 577 and 579 nm (yellow, two lines close enough that a homemade spectroscope may show them as one).

Safety. Never look at the Sun, either directly or through your spectroscope; to see sunlight's spectrum, aim the slit at white paper lit by the Sun, as NASA's instructions advise. Cut away from your fingers, and take special care with a craft knife or saw. Do not remove, handle or break fluorescent tubes or bulbs, which contain mercury; look only at lamps that are already fixed in place.

In a school lab the same comparison is sharper with gas discharge tubes and a transmission grating. The tubes run on a high-voltage supply, about 5000 volts for the hydrogen tube HyperPhysics describes, so the teacher sets them up and switches them on; never touch the terminals, and let a tube cool before it is moved. With a 600 lines/mm grating, d = 1/600 mm = 1.667 × 10^-6 m, and the red hydrogen line appears in first order at sinθ = λ/d = (656 × 10^-9 m)/(1.667 × 10^-6 m) = 0.394, so θ = 23.2°. Measuring that angle and working backwards is the classic way to measure a spectral line, and it is the experimental-design task in this lesson's activity.

Common misconceptions

• "An electron can orbit at any radius, as long as it moves at the right speed." In Bohr's model only radii r_n = n²a_B are allowed, the ones whose circumference holds a whole number of de Broglie wavelengths.
• "An atom can absorb a photon with a bit more energy than a gap and keep the extra." Below the ionization energy, only an exact match is absorbed; other photons pass straight through. A colliding electron, by contrast, can hand over part of its energy.
• "A negative energy means something has gone wrong." Zero is chosen for a free electron at rest far from the nucleus. Negative means bound, and the ionization energy is the positive amount needed to reach zero.
• "An element's absorption lines and emission lines come at different wavelengths." For the same pair of levels they are identical. A cool gas simply shows fewer absorption lines, only those that start from the levels its atoms occupy.
• "The ground-state electron is slowest, because it has the least energy." It is the fastest, at 2.19 × 10⁶ m/s. Its total energy is lowest because its potential energy is the most negative, which is why it is the hardest to remove.
• "Bohr's orbits show what atoms really look like." Bohr's model is historical. It fails for every atom with more than one electron, and modern theory keeps the energy levels but has no well-defined orbits.

Pulling it together

Three comparisons carry this lesson. The planetary atom and the Bohr atom share the Coulomb force, but Bohr allowed only orbits whose circumference holds a whole number of de Broglie wavelengths, and that single rule turns Coulomb's law and circular motion into hydrogen's levels, E_n = -13.6 eV/n², with the ground-state electron moving at 2.19 × 10⁶ m/s and bound by 13.6 eV. Emission and absorption are the same jumps run in opposite directions: a drop emits a photon whose energy equals the gap, and an atom absorbs only a photon that matches a gap from the level it is in, so cool hydrogen is invisible in absorption while warmer hydrogen shows the Balmer lines at 656, 486, 434 and 410 nm. And photons and colliding electrons both excite atoms, but a photon must match a gap exactly while an electron can hand over part of its energy, as Franck and Hertz's 4.9 eV steps in mercury showed. Every element has its own ladder of levels and so its own barcode of lines, bright or dark at the same wavelengths. The next lesson turns from the electrons to the nucleus, where a small amount of missing energy in beta decay led to a new particle.

Sources

1. OpenStax. (2022). 30.3 Bohr's theory of the hydrogen atom (the collapse of classical orbits; m_evr = nh/2π; E_n = -13.6 eV/n²; the Bohr radius; the Lyman, Balmer and Paschen series; the limits of the model). In College Physics 2e. openstax.org
2. OpenStax. (2022). 30.2 Discovery of the parts of the atom: Electrons and nuclei. In College Physics 2e. openstax.org
3. OpenStax. (2016). 6.5 De Broglie's matter waves (the standing-wave condition 2πr_n = nλ and Example 6.12). In University Physics Volume 3. openstax.org
4. OpenStax. (2022). 5.3 Spectroscopy in astronomy (Fraunhofer's dark lines; the three kinds of spectra; sodium and helium in the Sun). In Astronomy 2e. openstax.org
5. OpenStax. (2022). 5.5 Formation of spectral lines (absorption needs an exact match; re-emission in all directions; Balmer absorption needs atoms in level 2; ionized hydrogen). In Astronomy 2e. openstax.org
6. Sansonetti, J. E., & Martin, W. C. (2013). Handbook of basic atomic spectroscopic data (NIST Standard Reference Database 108): strong lines of hydrogen and of mercury. National Institute of Standards and Technology. physics.nist.gov (hydrogen); physics.nist.gov (mercury)
7. National Institute of Standards and Technology. (2024). CODATA 2022 values: Bohr radius; proton-electron mass ratio. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
8. Nave, R. (n.d.). Franck-Hertz experiment. HyperPhysics, Georgia State University. hyperphysics.gsu.edu
9. Nebraska Astronomy Applet Project. (n.d.). The hydrogen atom: Student guide (which photon energies excite or ionize a ground-state atom). University of Nebraska-Lincoln. astro.unl.edu
10. NASA STEREO Learning Center. (n.d.). Cereal box spectroscope. stereo-ssc.nascom.nasa.gov
11. Swiss Science Center Technorama. (n.d.). CD spectroscope. explorer.technorama.ch
12. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 15.2 and 15.3. apcentral.collegeboard.org

The decay that seemed to lose energy

Radioactive decay is the spontaneous change of a nucleus into one or more different nuclei, or into a lower energy state of the same nucleus. By the 1920s physicists had measured the energies of the particles that unstable nuclei throw out, and two kinds of decay told two different stories. The alpha particles from a given decay come out with sharply defined energies, and so do the gamma rays; both show that a nucleus, like an atom, has definite energy levels. The electrons from beta decay do not. In 1914 James Chadwick found that the electrons from a given beta decay emerge with a broad, continuous spread of energies, from almost nothing up to a maximum.

That should not happen if a nucleus simply turns into another nucleus plus an electron. The parent and the daughter nuclei each have a definite energy, so every decay releases the same energy, and, as this lesson shows, two products sharing a fixed energy must always split it the same way. Either some energy was being lost on the way out, or energy was not conserved in beta decay, or something else was leaving unseen. The argument over which lasted about twenty years, and the answer added a new particle to physics. This lesson follows the dispute, then uses what it settled to describe the kinds of decay the AP exam asks about (Topic 15.8) and the half-life rules that describe a sample of decaying nuclei (Topic 15.7).

Four ways for a nucleus to decay

Which kind of decay a nucleus undergoes depends on which isotope it is, and the exam does not expect you to memorize which isotope does what. Each kind has a symbol and a rule for what changes.

Three new particles appear in that table. A positron, written e⁺ or β⁺, is the antielectron: the same mass as an electron and the opposite charge. A neutrino, written ν, and its antiparticle the antineutrino, written ν̄, have no electric charge and a mass small enough to neglect. They interact with matter only through the weak force and gravity, so they pass through ordinary matter almost without a trace. The AP course does not ask how neutrinos and antineutrinos differ, or how the weak force works.

Every nuclear decay obeys three counting rules, which let you complete any decay equation:

• Nucleon number, the number of protons plus neutrons (the top number, A), is the same before and after.
• Charge is the same before and after. For a nucleus the bottom number is its charge in units of e; give the electron a bottom number of -1 and the positron +1.
• Lepton number is the same before and after. Electrons and neutrinos count +1, positrons and antineutrinos count -1, and nucleons and photons count 0.

Energy, momentum and angular momentum are conserved as well, and that is where the trouble with beta decay began.

Worked problem: completing decay equations

Givens. Write the complete equation for each decay: (a) carbon-14, ¹⁴₆C, by beta-minus decay; (b) sodium-22, ²²₁₁Na, by beta-plus decay; (c) uranium-238, ²³⁸₉₂U, by alpha decay; (d) an excited nickel-60 nucleus, ⁶⁰₂₈Ni*, which is what cobalt-60 most often leaves behind when it decays.

Diagram in words. For each decay, a box for the parent on the left, an arrow, and a box for each product on the right, every box labelled with its nucleon number, its charge and its lepton number.

Principle. Nucleon number, charge and lepton number are each the same on both sides, and the number of protons, Z, fixes the element.

Algebra.

1. (a) A neutron becomes a proton, so Z goes from 6 to 7, which is nitrogen: ¹⁴₆C → ¹⁴₇N + e^- + ν̄. Check: nucleons, 14 = 14 + 0 + 0; charge, 6 = 7 + (-1) + 0; lepton number, 0 = 0 + 1 + (-1).
2. (b) A proton becomes a neutron, so Z goes from 11 to 10, which is neon: ²²₁₁Na → ²²₁₀Ne + e⁺ + ν. Check: 22 = 22; 11 = 10 + 1; 0 = 0 + (-1) + 1.
3. (c) The alpha particle takes 2 protons and 2 neutrons, leaving Z = 90, thorium, with A = 234: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Check: 238 = 234 + 4; 92 = 90 + 2; no leptons on either side.
4. (d) Only the energy changes: ⁶⁰₂₈Ni* → ⁶⁰₂₈Ni + γ.

Sanity check. In (a) and (b) the element changes but A does not, as the table predicts for beta decay; in (c) both change; in (d) nothing changes but the energy. Every lepton created has a partner of opposite lepton number, so leptons appear in pairs: an electron with an antineutrino, or a positron with a neutrino.

The variant that flips the result. Every decay above except gamma changed the element. Now let a nucleus undergo one alpha decay followed by two beta-minus decays. Z changes by -2, then +1, then +1, and ends where it started; A changes by -4, then 0, then 0. Three decays later the nucleus is the same element again, now a lighter isotope of it, four nucleons lighter.

Worked problem: why two products give a sharp energy

Givens. A plutonium-239 nucleus at rest decays by alpha emission, ²³⁹₉₄Pu → ²³⁵₉₂U + ⁴₂He, releasing 5.25 MeV as kinetic energy of the products. OpenStax finds this energy from the masses of the atoms, and Lesson 31 shows how. Take the masses of the products in the ratio of their nucleon numbers, 235 to 4. How is the energy shared?

Diagram in words. Before: one nucleus at rest. After: the alpha particle flying off to the right with momentum p, and the uranium nucleus recoiling to the left with momentum of the same size.

Principle. Momentum is conserved, so the two products have equal and opposite momenta. Kinetic energy can be written K = p²/(2m), so with equal momenta the kinetic energies are in the inverse ratio of the masses. Energy is conserved, so the two kinetic energies add up to 5.25 MeV.

Algebra with units. K_α/K_U = m_U/m_α = 235/4. So K_α = (5.25 MeV)(235/239) = 5.16 MeV, and K_U = (5.25 MeV)(4/239) = 0.088 MeV.

Sanity check. The light alpha particle takes about 98 percent of the energy and the heavy nucleus recoils slowly, like a rifle firing a bullet; OpenStax notes that the recoil of the uranium nucleus carries much less energy, as momentum conservation requires. The alpha particle's speed, from K = (1/2)mv² with m = 6.64 × 10^-27 kg, is about 1.6 × 10⁷ m/s, about 5 percent of the speed of light, so the non-relativistic formula is fine. Most important, nothing in the algebra was left to chance. Two products, a fixed total energy and momentum conservation allow only one answer, so every such decay produces an alpha particle with exactly the same energy: a sharp line in the alpha spectrum.

The variant that flips the result. Apply the same reasoning to the beta-minus decay of carbon-14 as if it had only two products, ¹⁴₆C → ¹⁴₇N + e^-. From NIST's atomic masses, 14.003 241 988 u for carbon-14 and 14.003 074 004 u for nitrogen-14, the energy released is (0.000 167 984 u)(931.5 MeV/u) = 0.156 MeV. (For beta-minus decay the masses of the neutral atoms can be compared directly, because the emitted electron's mass is already counted in the daughter atom.) The nitrogen nucleus is about 25 000 times as massive as the electron, so by the same argument the electron would take essentially all of the energy: every electron would leave with 0.156 MeV. Instead, the electrons from carbon-14, like those from every beta emitter, leave with any energy from nearly zero up to 0.156 MeV. The sharp line has become a smooth spread, and the same two conservation laws that explain the alpha line say that a beta decay with only two products cannot produce it.

The dispute: where did the energy go?

Three explanations competed, and each made a prediction that could be tested.

Position 1: the energy is lost on the way out. Some physicists, Lise Meitner among them, thought every electron left the nucleus with the full energy and then lost varying amounts as it struggled out through the strong electric field around the nucleus, radiating the difference. If so, the lost energy still exists as radiation, and a thick enough absorber around the sample should catch all of it.

In 1927 C. D. Ellis and William Wooster at the Cavendish Laboratory in Cambridge tested this with a calorimeter: a thick-walled container around a beta source, radium E, an isotope of bismuth, that turned everything it absorbed into heat. If position 1 were right, the heat per decay would equal the maximum electron energy. It equalled the average electron energy instead. Meitner, repeating the measurement with W. Orthmann, confirmed the result and showed that no gamma rays were involved. The missing energy was not being absorbed by the walls, so position 1 failed.

Position 2: energy is conserved only on average. Niels Bohr was willing to give up the conservation of energy in individual nuclear processes, keeping it only as a statistical law, in the way the second law of thermodynamics is statistical. On this view nothing was escaping the calorimeter; energy simply was not conserved decay by decay.

Position 3: an unseen particle carries it off. On 4 December 1930 Wolfgang Pauli sent an open letter to a physics meeting in Tübingen. It opened "Dear radioactive ladies and gentlemen," and offered what Pauli called a desperate remedy: a very light, electrically neutral particle, emitted together with the electron in such a way that the sum of their energies is always the same. A neutral particle that barely interacts would pass straight through the calorimeter walls, which explains Ellis and Wooster's result.

Pauli also pointed to the observation that would decide between him and Bohr: whether the beta spectrum has a sharp upper limit. If energy is conserved only on average, some electrons could come out with more than the energy difference between the parent and daughter nuclei, and the spectrum would tail away with no definite end. If an unseen partner shares a fixed energy, no electron can ever get more than all of it, and the spectrum must stop sharply at exactly the energy released. In 1933 Ellis and Mott argued that the spectrum does end sharply, at the energy difference between parent and daughter, and noted that if the energy simply vanished it would not be conserved even on average, since the calorimeter recorded only the average electron energy. Beta spectra are now known to be continuous with an end point equal to the energy released, Q.

The theory came next. Enrico Fermi built Pauli's particle into a theory of beta decay in 1934, in which the electron and the new particle are both created at the moment of decay, and he named it the neutrino, the little neutral one; the name neutron had by then gone to the heavy particle James Chadwick found in 1932. Detection took another two decades. Hans Bethe and Rudolf Peierls estimated that stopping neutrinos from a radioactive source efficiently would need a target several light years thick. Frederick Reines and Clyde Cowan used a nuclear reactor instead, where about 10¹³ antineutrinos crossed each square centimetre every second, and looked for the reaction antineutrino + proton → neutron + positron in a few hundred litres of water. (Check the lepton number: -1 + 0 = 0 + (-1).) After a first trial at the Hanford reactor, they detected the antineutrino at the Savannah River reactor, recording a few events per hour, and published the result in 1956. Reines shared the 1995 Nobel Prize in Physics for the detection of the neutrino; Cowan had died in 1974.

Worth holding on to: when a conservation law seems to fail, the first suspect is something you are not detecting. Pauli kept energy and momentum conservation and predicted a particle; a continuous beta spectrum with a sharp end point is the evidence that a third, unseen product shares the energy, and the detection of the neutrino a quarter of a century later confirmed it.

Half-life: no one can say when a nucleus will decay

Now from the single nucleus to the sample. The moment at which a particular nucleus decays is, in the CED's word, indeterminable; all that can be given is a probability. What can be predicted is the behaviour of a large number of nuclei. The half-life, t_1/2, is the time it takes for half of the radioactive nuclei in a sample to decay. Each nucleus has a 50 percent chance of surviving for one half-life, and, as OpenStax stresses, that chance is the same no matter when you start counting.

After one half-life half the original nuclei remain; after two, a quarter; after three, an eighth. In general,

N = N0(1/2)t/t1/2 = N0e-λt, with λ = ln 2/t1/2 ≈ 0.693/t1/2

Here N₀ is the number of nuclei at t = 0, and λ is the decay constant: the larger it is, the faster the sample decays and the shorter its half-life. The same Greek letter is used for wavelength, and the context tells you which is meant. Taking the natural logarithm of both sides gives the form the CED lists as a derived equation, ln(N/N₀) = -λt, which is the key to finding an age or a half-life from data.

The values come from OpenStax's table of selected isotopes. The CED notes that half-lives range from fractions of a second to billions of years, and the exam gives you any half-life you need.

A Geiger counter counts decays, not nuclei. The activity of a sample, its number of decays per second, is R = λN, measured in becquerels (1 Bq is one decay per second). Because R is proportional to N, the activity also halves every half-life, so a graph of count rate against time shows the same half-life as a graph of N.

Worked problem: dating a piece of charcoal

Givens. Charcoal from an ancient hearth contains 25.0 percent of the carbon-14 found in the same mass of carbon from living wood. Carbon-14 has a half-life of 5730 years. A living tree keeps exchanging carbon with its surroundings; when it dies the exchange stops, and its carbon-14 is no longer replenished as it decays. How long ago did the tree die, and is the hearth more than 10 000 years old?

Diagram in words. A decay curve of N/N₀ against time, starting at 1, falling to 0.5 at 5730 years and to 0.25 at 11 460 years, with a horizontal line at 0.25 drawn across to meet it.

Principle. N/N₀ = (1/2)^t/t_1/2, or equivalently ln(N/N₀) = -λt with λ = 0.693/t_1/2.

Algebra with units. 0.250 = (1/2)², so t = 2t_1/2 = 2(5730 y) = 11 460 years. With logarithms, which work for any fraction: λ = 0.693/(5730 y) = 1.21 × 10^-4 per year, and t = -ln(0.250)/λ = 1.386/(1.21 × 10^-4 per year) = 11 500 years, the same to the three figures the data justify. So yes, the hearth is more than 10 000 years old.

Sanity check. Two half-lives leave a quarter, so the answer had to be two half-lives. OpenStax works a famous case the same way: samples of the Shroud of Turin held 92 percent of the carbon-14 in living tissue, which gives an age of about 690 years, a date near AD 1300.

The variant that flips the result. Suppose instead that the charcoal holds 30.0 percent of the living level. It is tempting to call that a little less than two half-lives and still guess more than 10 000 years. Do the arithmetic: t = -ln(0.300)/λ = 1.204/(1.21 × 10^-4 per year) = 9950 years. The answer flips: this hearth is younger than 10 000 years. Exponential decay is not a straight line, so fractions between the half-life marks need the logarithm, not a guess.

Graphs: a curve that becomes a straight line

A graph of N against t is a falling exponential: it drops by half in every half-life, steeply at first and then more and more slowly. The smooth model curve approaches zero without ever reaching it; a real sample, being a finite number of nuclei, eventually runs out, at a moment no one can predict. To test a model and pull a constant out of data, the exam's laboratory questions favour a straight line, and ln(N/N₀) = -λt supplies one. Plot ln N against t and the points fall on a line with slope -λ and intercept ln N₀. A steeper line means a larger decay constant and a shorter half-life, t_1/2 = 0.693/λ. Count rates work just as well as numbers of nuclei, because the activity is proportional to N.

Try it: radioactive dice

You cannot keep radioactive sources at home, but you can model them. The College Board's course description suggests this activity for a class: 200 dice in a box with a lid; shake, open, remove every die showing a 1, and repeat with the dice that are left. Each die, like each nucleus, has the same fixed chance of "decaying" on every throw, one in six, however long it has already survived.

1. Gather as many dice as you can, at least 50; 100 or more gives smoother results. Coins work too: remove every coin that lands heads.
2. Throw all the remaining dice, remove the ones, and record how many are left. Repeat until none or only a few remain.
3. Plot the number left, N, against the number of throws. Then plot ln N against the number of throws and draw the best-fit line.
4. From the slope, find the decay constant per throw and the half-life in throws, and compare them with the prediction in the activity answers.

The CED's version of the activity expects a half-life of about 3.8 turns. If your school has a Geiger-Müller counter and sealed radioactive sources, your teacher may demonstrate real count rates; only the teacher handles the sources, following the school's rules for radioactive materials.

Common misconceptions

• "Beta electrons were sitting inside the nucleus, waiting to get out." They are created at the moment of decay, as a neutron changes into a proton. Fermi's theory, built on Pauli's neutrino, made this the accepted picture.
• "After two half-lives the sample is all gone." Half of a half remains: a quarter. After ten half-lives about a thousandth remains, since (1/2)¹⁰ = 1/1024.
• "A nucleus that has survived a long time is due to decay." Its chance of decaying in the next half-life is 50 percent, whatever its age.
• "A bigger sample has a longer half-life." The half-life belongs to the isotope. A bigger sample has more decays per second but halves in the same time.
• "Gamma decay turns one element into another." A gamma photon carries away energy only; Z and A are unchanged.
• "The neutrino was just a trick to save a theory." It began as a hypothesis, but it made a testable prediction, a sharp end to the beta spectrum, and it was detected directly at a reactor in 1956.

What you now know

Alpha particles and gamma rays from a given decay have sharp energies because two products sharing a fixed energy must split it one way; for plutonium-239 the alpha particle takes 5.16 MeV of the 5.25 MeV released. Beta electrons come out with every energy up to a sharp end point, which Chadwick found in 1914 and which no two-product decay can produce. Of the three explanations, energy lost on the way out failed Ellis and Wooster's calorimeter test in 1927, and energy conserved only on average failed against the sharp end point; Pauli's neutral, penetrating particle survived, Fermi built it into a theory in 1934, and Reines and Cowan detected it at a reactor in 1956. In every decay, alpha, beta-minus, beta-plus or gamma, nucleon number, charge and lepton number are conserved, which is all you need to complete a decay equation. For a sample, only probabilities hold for single nuclei, but large numbers follow N = N₀e^-λt with λ = 0.693/t_1/2; a graph of ln N against time is a straight line of slope -λ, and the same equation dates charcoal, or a shroud, from the carbon-14 that is left. The next lesson asks where the energy released in a decay comes from, and finds it in the masses.

Sources

1. OpenStax. (2022). 31.4 Nuclear decay and conservation laws (alpha, beta-minus, beta-plus and gamma decay; conservation of nucleon number, charge and electron family number; the plutonium-239 and cobalt-60 examples; Pauli and Fermi). In College Physics 2e. openstax.org
2. OpenStax. (2022). 31.5 Half-life and activity (N = N₀e^-λt; carbon-14 dating and the Shroud of Turin example; activity in becquerels). In College Physics 2e. openstax.org
3. OpenStax. (2022). Appendix B: Selected radioactive isotopes. In College Physics 2e. openstax.org
4. Brown, L. M. (1978). The idea of the neutrino (Chadwick's 1914 continuous spectrum; Ellis and Wooster's calorimeter; Bohr's statistical view; Ellis and Mott; Pauli's letter of 4 December 1930 in translation). Physics Today, 31(9), 23. physicstoday.aip.org
5. Bilenky, S. M. (2012). Neutrino. History of a unique particle (arXiv:1210.3065; the 1927 calorimeter date, the end point equal to Q, and the Reines and Cowan experiments). arxiv.org
6. The Royal Swedish Academy of Sciences. (1995). The Nobel Prize in Physics 1995 [Press release] (Pauli's proposal; Bethe and Peierls; the Reines and Cowan reactor experiment). nobelprize.org
7. Coursey, J. S., Schwab, D. J., Tsai, J. J., & Dragoset, R. A. (n.d.). Atomic weights and isotopic compositions: relative atomic masses of carbon-14 and nitrogen-14. National Institute of Standards and Technology. physics.nist.gov (carbon); physics.nist.gov (nitrogen)
8. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topics 15.7 and 15.8, and the radioactive dice activity among the Unit 15 sample activities. apcentral.collegeboard.org

Where the energy comes from

Lesson 30 quoted energies without saying where they came from: 5.25 MeV released when plutonium-239 emits an alpha particle, 0.156 MeV when carbon-14 beta decays. Both came from masses. In every nuclear reaction that releases energy, the products have slightly less mass than the starting particles, and the missing mass reappears as energy according to Einstein's

E = mc2

This is mass-energy equivalence: mass is a form of energy, and a change in the energy of a system shows up as a change in its mass. The exchange rate is enormous. OpenStax works out that 1.00 g of mass is equivalent to 9.00 × 10¹³ J. The same rule holds in ordinary life, but there the changes are hopelessly small: OpenStax's example of charging a car battery shows that you would need to measure the battery's mass to a billionth of a percent to notice the change. In nuclear processes the energy per particle is so large that the mass change is easy to measure.

The CED (Topic 15.7) lists the rules every nuclear reaction obeys: nucleon number is conserved, and so are energy, counting the rest energy mc² of every particle, and momentum. The energy released appears as kinetic energy of the products or as photons. This lesson turns those rules into a procedure and uses it on the three processes that power the Sun, nuclear reactors and the decays of Lesson 30.

The strong force and the binding energy

A nucleus packs positively charged protons into a few femtometres (1 fm = 10^-15 m), where their electric repulsion is enormous. What holds it together is the strong force, which acts between nucleons, protons and neutrons alike. OpenStax describes it as about 100 times stronger than the electric repulsion at nuclear distances, but so short-ranged that it falls to zero when nucleons are only a few femtometres apart, so each nucleon feels the pull of its near neighbours only.

Because the nucleons are bound, you would have to do work to pull a nucleus apart into separate protons and neutrons. That energy is the nucleus's binding energy, BE. By mass-energy equivalence, a bound nucleus must have less mass than its separate parts, by exactly BE/c². That missing mass is called the mass defect. Using the masses of neutral atoms, which keeps the electrons balanced,

BE = [Zm(1H) + Nmn - m(AX)]c2

where m(¹H) is the mass of a hydrogen atom, m_n the mass of a neutron, Z the number of protons and N the number of neutrons. Dividing by A gives the binding energy per nucleon, the measure of how tightly a nucleus is held together.

The procedure: from masses to energy

Masses in nuclear physics are given in unified atomic mass units, u, where 1 u = 1.6605 × 10^-27 kg. The energy equivalent of 1 u is 931.5 MeV (CODATA 2022 gives 931.494 103 72 MeV), which turns every calculation into a single multiplication. The steps are the same for any reaction:

1. Write the reaction and check that nucleon number and charge balance.
2. Add up the masses before and the masses after, in u. Use the masses of neutral atoms; this works whenever the number of electrons in the atoms is the same on both sides, as it is in every reaction in this lesson.
3. Find the mass difference, Δm = (mass before) - (mass after).
4. Convert: Q = (Δm)(931.5 MeV/u), or Q = Δmc² in joules with Δm in kilograms.
5. Read the sign. If Q is positive, the reaction releases Q as kinetic energy of the products or as photons. If Q is negative, it cannot happen on its own; at least |Q| must be supplied.
6. If you need to know which product gets the energy, share it out with conservation of momentum, as in Lesson 30.

Worked problem: what holds helium-4 together

Givens. From NIST's atomic masses: helium-4, m(⁴He) = 4.002 603 u; a hydrogen atom, m(¹H) = 1.007 825 u; a neutron, m_n = 1.008 665 u. Find the binding energy of helium-4 and its binding energy per nucleon.

Diagram in words. On the left, one helium-4 atom. On the right, the same nucleons pulled far apart: two hydrogen atoms and two free neutrons. An arrow from left to right is labelled "energy supplied: BE".

Principle. Conservation of energy with mass-energy equivalence: the separated parts have more mass than the helium atom by BE/c². Two hydrogen atoms carry the two electrons that the helium atom has, so the electrons balance.

Algebra with units. Separated parts: 2(1.007 825 u) + 2(1.008 665 u) = 4.032 980 u. Mass defect: Δm = 4.032 980 u - 4.002 603 u = 0.030 377 u. Binding energy: BE = (0.030 377 u)(931.5 MeV/u) = 28.3 MeV. Per nucleon: 28.3 MeV/4 = 7.07 MeV.

Sanity check. OpenStax gets the same 28.3 MeV and 7.07 MeV per nucleon. The mass defect is 0.75 percent of the mass, and the binding energy is about two million times the 13.6 eV that binds hydrogen's electron in Lesson 29, which is why nuclear energies are measured in MeV and atomic ones in eV.

The variant that flips the result. Repeat for the deuteron, m(²H) = 2.014 102 u: Δm = 1.007 825 u + 1.008 665 u - 2.014 102 u = 0.002 388 u, so BE = 2.22 MeV, only 1.11 MeV per nucleon. Now let two deuterons combine into helium-4. Taking helium apart costs 28.3 MeV; taking two deuterons apart costs only 2(2.22 MeV) = 4.45 MeV. So building helium-4 from two deuterons releases 28.3 MeV - 4.45 MeV = 23.85 MeV, the value OpenStax lists for that fusion reaction. Energy had to be supplied to separate helium; energy comes out when loosely bound nuclei combine into a tightly bound one. In general, Q equals the binding energy of the products minus the binding energy of the starting nuclei.

The binding-energy curve

Work out the binding energy per nucleon for nucleus after nucleus and plot it against A, and a single curve appears. It rises steeply through the lightest nuclei, with helium-4 sitting unusually high, levels off at about 8 MeV per nucleon, peaks near A = 60, around iron, where nuclei are most tightly bound, and falls slowly towards the heaviest nuclei. The values below were computed with the procedure from NIST's atomic masses.

The shape follows from the two forces. In a light nucleus, each nucleon has few neighbours to bind it, so adding nucleons raises the binding energy per nucleon. In a heavy nucleus, every nucleon already has its full set of neighbours, while the electric repulsion, which reaches right across the nucleus, keeps growing with the number of protons, so the binding energy per nucleon slowly falls.

The point: energy is released whenever nucleons end up more tightly bound than they started, that is, whenever a reaction moves them up the binding-energy curve towards iron. Fusing light nuclei moves them up from the left; splitting heavy nuclei moves them up from the right. Going the other way, fusing nuclei beyond iron or splitting nuclei near it, needs an energy input.

Worked problem: fusing deuterium and tritium

Givens. The fusion reaction planned for fusion reactors is ²H + ³H → ⁴He + n. Atomic masses: m(²H) = 2.014 102 u, m(³H) = 3.016 049 u, m(⁴He) = 4.002 603 u, m_n = 1.008 665 u. Find the energy released and how it is shared, taking the starting nuclei as nearly at rest.

Diagram in words. Two small nuclei approach and merge; afterwards a helium-4 nucleus moves off one way and a neutron the opposite way, the neutron's arrow four times as long as the helium's.

Principle. The six-step procedure for Q, then momentum conservation to share it: the two products leave with equal and opposite momenta, so their kinetic energies are in the inverse ratio of their masses.

Algebra with units. Check the balance: nucleons, 2 + 3 = 4 + 1; charge, 1 + 1 = 2 + 0. Before: 2.014 102 u + 3.016 049 u = 5.030 151 u. After: 4.002 603 u + 1.008 665 u = 5.011 268 u. Δm = 0.018 883 u, so Q = (0.018 883 u)(931.5 MeV/u) = 17.59 MeV. Shared in the inverse ratio of the masses, about 4 to 1: the neutron gets (4/5)(17.59 MeV) = 14.1 MeV and the helium nucleus (1/5)(17.59 MeV) = 3.5 MeV.

Sanity check. OpenStax lists 17.59 MeV for this reaction. The energy per nucleon released, 17.59 MeV/5 = 3.5 MeV, matches the jump from the weakly bound deuterium and tritium (1.11 and 2.83 MeV per nucleon) to helium-4 (7.07 MeV per nucleon). Q is positive, so the reaction releases energy.

The variant that flips the result. In Lesson 30's alpha decay of plutonium, the alpha particle took 98 percent of the energy. Here the same particle, a helium-4 nucleus, takes only 20 percent. Nothing about helium changed: in the decay it was the light partner of a uranium nucleus, and here it is the heavy partner of a neutron. Momentum conservation always hands most of the energy to the lighter product.

Fusion in the Sun

The Sun produces its energy by fusing hydrogen nuclei. In the chain of reactions OpenStax describes, the overall effect is to turn four hydrogen nuclei into one helium-4 nucleus, releasing 26.7 MeV (a figure that includes the annihilation of the positrons made along the way). Using atomic masses, 4(1.007 825 u) - 4.002 603 u = 0.028 697 u, which is 0.7 percent of the starting mass and 26.7 MeV.

Fusion is hard to start because nuclei are positively charged: they repel, and they must get within a few femtometres before the strong force can take over. OpenStax notes that pushing nuclei into contact against this repulsion would classically need temperatures of about 10⁸ K, more than the Sun's core temperature; quantum tunnelling lets the Sun fuse hydrogen anyway. The rate is set by the Sun's power. NASA gives its luminosity as 3.828 × 10²⁶ W, so the Sun turns mass into energy at Δm/Δt = P/c² = (3.828 × 10²⁶ J/s)/(3.00 × 10⁸ m/s)² = 4.26 × 10⁹ kg every second, the same mass conversion rate NASA's Sun fact sheet lists. That is over four million tonnes a second, yet with a mass of 1.99 × 10³⁰ kg the Sun loses less than a ten-trillionth of its mass a year.

Worked problem: splitting uranium-235, and why iron will not split

Givens. A slow neutron is absorbed by uranium-235, and the nucleus splits: n + ²³⁵U → ¹⁴²Ba + ⁹¹Kr + 3n, the example OpenStax uses. Atomic masses: m(²³⁵U) = 235.043 930 u, m(¹⁴²Ba) = 141.916 432 u, m(⁹¹Kr) = 90.923 806 u, m_n = 1.008 665 u. Find the energy released.

Diagram in words. A neutron arrives at a large nucleus; the nucleus stretches into a dumbbell and splits into two medium-sized fragments flying apart, with three new neutrons leaving in different directions.

Principle. The six-step procedure, with the masses of neutral atoms: uranium has 92 electrons, and barium (56) and krypton (36) together have 92, so the electrons balance.

Algebra with units. Check: nucleons, 1 + 235 = 142 + 91 + 3 = 236; charge, 92 = 56 + 36. Before: 1.008 665 u + 235.043 930 u = 236.052 595 u. After: 141.916 432 u + 90.923 806 u + 3(1.008 665 u) = 235.866 233 u. Δm = 0.186 362 u, so Q = (0.186 362 u)(931.5 MeV/u) = 174 MeV.

Sanity check. The answer is positive and huge, about ten times the energy of a D-T fusion, though spread over 236 nucleons rather than 5. In binding-energy terms, the nucleons move from 7.59 MeV per nucleon in uranium-235 to 8.31 MeV in barium-142 and 8.54 MeV in krypton-91 (both computed from NIST's masses), up the curve towards iron. OpenStax uses about 200 MeV as an average for the fission of uranium-235, since the nucleus can split in many different ways, and finds that 1.00 kg of uranium-235 fissioned this way releases about 8.21 × 10¹³ J, equivalent to about 14 000 barrels of crude oil.

The variant that flips the result. Try to split iron-56 into two aluminium-28 nuclei, ⁵⁶Fe → ²⁸Al + ²⁸Al. Before: 55.934 936 u. After: 2(27.981 910 u) = 55.963 820 u. Now Δm = -0.028 884 u and Q = -26.9 MeV. The products have more mass than the iron nucleus, so this fission can never happen on its own: at least 26.9 MeV would have to be supplied. Iron sits at the top of the binding-energy curve (8.79 MeV per nucleon), and its fragments are less tightly bound (8.31 MeV per nucleon for aluminium-28), so splitting it moves the nucleons down the curve, not up.

Spontaneous and induced fission, and the chain reaction

Whether a heavy nucleus splits on its own depends on its binding energy. Some nuclei undergo spontaneous fission, though OpenStax notes that this is usually not their most common way to decay. Fission can also be induced: a nucleus absorbs a neutron, and the energy the neutron deposits sets the nucleus vibrating and stretching, like a liquid drop, until it pulls apart. That is the energy input the CED refers to when it says fission may occur spontaneously or may require an energy input.

Each fission of uranium-235 releases several neutrons, three in the example above, and each can trigger another fission. A self-sustaining chain reaction is possible provided that, on average, the neutrons from one fission cause at least one more. In a small lump, too many neutrons escape through the surface before they are absorbed, so there is a minimum amount needed for a self-sustaining chain reaction, called the critical mass. In a reactor, the chain reaction is controlled so that each fission leads, on average, to just one more.

So the CED's two definitions, read against the curve, make sense together. Nuclear fusion is two or more smaller nuclei combining into a larger nucleus, with other subatomic particles; it releases energy for light nuclei. Nuclear fission is a nucleus splitting into two or more smaller nuclei, with other subatomic particles; it releases energy for heavy nuclei.

Try it: a binding-energy curve and a domino chain

No nuclear experiment is safe at home, but the two big ideas of this lesson can be modelled. The activity has you build the binding-energy curve from NIST's measured masses, which is a graphing task of exactly the kind the exam sets. For the chain reaction, a set of dominoes will do.

1. Stand 15 to 20 dominoes in a single line and knock over the first: one domino falls at a time, steadily.
2. Now build a branching pattern in which each domino knocks over two: one in the first row, two in the second, four in the third, and so on for four or five rows. Knock over the first.
3. Count how many dominoes fall in each row, and compare how the falling spreads in the two set-ups.

The line is like a chain reaction in which each fission causes exactly one more; the branching pattern is like one in which each causes two, and the number falling doubles row by row. The model leaves out a great deal: neutrons that escape or are absorbed without causing fission, the energy released, and the time between steps. Keep fingers clear of falling dominoes, and nothing more is needed for safety.

Common misconceptions

• "Nuclear reactions destroy nucleons to make energy." Nucleon number is conserved in every reaction. The mass that disappears is the change in binding energy, as the nucleons end up more tightly bound.
• "E = mc² only applies to nuclear reactions." It applies to every change in energy, but in chemical and everyday changes the mass difference is far too small to measure.
• "Binding energy is energy stored in a nucleus, released when it breaks apart." It is the energy you must supply to take a nucleus apart. A more tightly bound nucleus has less mass, and energy is released when nucleons become more tightly bound, not less.
• "Fusion always releases energy and fission always costs energy." It depends on which side of the iron peak you are. Fusing light nuclei and splitting heavy ones release energy; splitting iron or fusing heavy nuclei costs it.
• "Fusion is easy, because the strong force is so strong." The strong force acts only within a few femtometres. The nuclei first have to get that close against their electric repulsion, which is why fusion needs extreme temperatures.
• "Any lump of uranium will run a chain reaction." On average each fission must cause at least one more, and in a lump smaller than the critical mass too many neutrons escape.

The takeaway

Every nuclear reaction conserves nucleon number, charge, energy and momentum, and mass counts as energy through E = mc², with 1 u equivalent to 931.5 MeV. The procedure is always the same: balance the reaction, compare the masses before and after, multiply the difference by 931.5 MeV/u, read the sign, and share the energy with momentum conservation. The strong force binds nucleons over a few femtometres, and the binding energy, the energy needed to take a nucleus apart, is the mass defect times c²: 28.3 MeV, or 7.07 MeV per nucleon, for helium-4. Binding energy per nucleon peaks near iron, so fusing light nuclei, such as deuterium and tritium (17.59 MeV, mostly to the neutron), and splitting heavy nuclei, such as uranium-235 (174 MeV for the split worked here), both release energy, while splitting iron would cost 26.9 MeV. The Sun fuses hydrogen into helium and converts 4.26 × 10⁹ kg of mass into energy every second; a reactor sustains a fission chain reaction in which each fission leads, on average, to one more. The last lesson of the course turns from physics to the exam itself.

Sources

1. OpenStax. (2022). 31.6 Binding energy (BE = [Zm(¹H) + Nm_n - m(^AX)]c²; helium-4 at 28.3 MeV and 7.07 MeV per nucleon; the peak near A = 60). In College Physics 2e. openstax.org
2. OpenStax. (2022). 31.3 Substructure of the nucleus (the range and strength of the nuclear force; 1 u = 931.5 MeV/c²). In College Physics 2e. openstax.org
3. OpenStax. (2022). 32.5 Fusion (the proton-proton cycle and its 26.7 MeV; the D-T reaction at 17.59 MeV; deuterium fusion at 23.85 MeV; the Coulomb barrier and tunnelling). In College Physics 2e. openstax.org
4. OpenStax. (2022). 32.6 Fission (the uranium-235 example; spontaneous and induced fission; chain reactions and critical mass; energy per kilogram). In College Physics 2e. openstax.org
5. OpenStax. (2022). 28.6 Relativistic energy (rest energy E₀ = mc²; the 1.00 g example; the car battery). In College Physics 2e. openstax.org
6. Coursey, J. S., Schwab, D. J., Tsai, J. J., & Dragoset, R. A. (n.d.). Atomic weights and isotopic compositions (relative atomic masses of the isotopes used here). National Institute of Standards and Technology. nist.gov
7. National Institute of Standards and Technology. (2024). CODATA 2022 values: neutron mass in u; atomic mass constant energy equivalent in MeV. NIST Reference on Constants, Units and Uncertainty. physics.nist.gov
8. NASA Space Science Data Coordinated Archive. (n.d.). Sun fact sheet (mass, luminosity and mass conversion rate). nssdc.gsfc.nasa.gov
9. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Topic 15.7 (Fission, fusion, and nuclear decay). apcentral.collegeboard.org

One exam, two sections, four forms

Everything in the previous 31 lessons is examined in a single sitting of three hours. The College Board lists the 2027 exam for Thursday 6 May 2027, and from that exam on the format is:

The exam is hybrid digital. You answer the multiple-choice questions in the Bluebook app and read the free-response questions there, but you write your free-response answers by hand in a paper booklet. A four-function, scientific or graphing calculator is allowed in both sections, and reference materials, the equation tables, are provided. The CED's 2026 clarifications record the change from 40 multiple-choice questions in 80 minutes and from 100 minutes of free response. In the multiple-choice section, Units 9, 10 and 11 carry 15 to 18 percent each and Units 12 to 15 carry 12 to 15 percent each. Science Practice 1, creating representations, is not assessed in the multiple-choice section; all three practices are assessed in free response.

The four free-response questions are always the same four types, in the same order, and each is a form with its own rules:

That is 40 points in 95 minutes. The suggested times add up to between 85 and 105 minutes, so there is no slack for a question that goes badly: decide in advance roughly when to move on.

The task verbs are instructions

Each part of a free-response question starts with a task verb, and the CED defines them. Reading the verb tells you what will earn credit. In brief:

• Calculate: work to a final answer, showing the expression, the substituted numbers, the units and sensible significant figures.
• Derive: start from a fundamental law or relationship and show the mathematical steps to the result.
• Justify: give reasoning in words, beyond the equations, that supports or defends a claim.
• Sketch: show the relationships and key features of a graph or model; an exact scale is not needed. Plot: place data points using a given or stated scale and units.
• Describe: give the relevant characteristics. Indicate: state the information, with no explanation needed. Determine: reach a conclusion by reasoning, observation or calculation.
• Estimate: calculate roughly, without needing to show steps. Verify: confirm that the conditions of a law or test are met, or that data support a hypothesis.

Most lost points come from answering a different verb from the one asked:

• a formula quoted where a derivation was asked for;
• a number without units where a calculation was asked for;
• a claim restated where a justification was asked for.

The rest of this lesson debugs four answers, one of each type, written for this course to show those failures. They are not College Board questions.

Bug hunt 1: a Mathematical Routines answer that forgot the battery

The task. A battery of emf 9.0 V and internal resistance 1.0 Ω is connected to a resistor R₁ = 8.0 Ω.

• (a) Derive an expression for the potential difference V_T across the battery's terminals, in terms of ε, r and R₁.
• (b) Calculate V_T.
• (c) A second 8.0 Ω resistor is then connected in parallel with R₁. Does V_T increase, decrease or stay the same? Justify your claim.

A student wrote:

(a) V = IR, so V_T = ε.

(b) V_T = 9.

(c) It stays the same, because resistors in parallel all get the full battery voltage.

The bugs.

1. Part (a) is not a derivation. It quotes Ohm's law and jumps to an answer that ignores r and does not even contain R₁. A derivation starts from a fundamental law, here Kirchhoff's loop rule, and shows the steps.
2. Part (b) has no unit, and it inherits the error from (a).
3. Part (c) states a rule instead of reasoning. Parallel branches do share one potential difference, but that potential difference is V_T, the very quantity in question, and it is not fixed at ε when the battery has internal resistance.

The fix, set out in full.

Givens. ε = 9.0 V, r = 1.0 Ω and R₁ = 8.0 Ω; later R₂ = 8.0 Ω in parallel with R₁.

Diagram in words. A single loop. The battery is drawn as an ideal emf with a 1.0 Ω resistor in series, both inside a dashed box, joined by wires to R₁. A voltmeter across the battery's terminals reads V_T. For part (c), R₂ is added as a second branch beside R₁.

Principle. Kirchhoff's loop rule, which is conservation of energy: around a closed loop the changes in potential add to zero, so ε - Ir - IR₁ = 0. The terminal voltage is the emf minus the drop across the internal resistance, V_T = ε - Ir.

Algebra with units.

• (a) From the loop rule, I = ε/(R₁ + r), so V_T = ε - εr/(R₁ + r) = εR₁/(R₁ + r).
• (b) V_T = (9.0 V)(8.0 Ω)/(8.0 Ω + 1.0 Ω) = 8.0 V, with I = 1.0 A.
• (c) Two 8.0 Ω resistors in parallel give R_eq = 4.0 Ω, so I = (9.0 V)/(4.0 Ω + 1.0 Ω) = 1.8 A and V_T = 9.0 V - (1.8 A)(1.0 Ω) = 7.2 V. V_T decreases.

Sanity check. The expression behaves sensibly at the limits. As R₁ becomes very large, no current flows and V_T approaches ε. As R₁ approaches zero, a short circuit, V_T approaches zero. Both sides are in volts. Here is the justification for (c), written as the step-by-step chain the MR question rewards:

1. Adding a parallel branch lowers the equivalent resistance of the external circuit.
2. With the same emf, a lower total resistance means a larger current.
3. A larger current means a larger potential drop Ir inside the battery.
4. So less of the emf is left across the terminals, and V_T falls.

The variant that flips the result. Make the battery ideal, r = 0. Then V_T = εR₁/R₁ = ε = 9.0 V whatever is connected, and the student's claim in (c), that V_T stays the same, becomes correct. It was wrong only because the internal resistance was not zero: a justification has to use the information given, not a rule remembered from ideal circuits.

Bug hunt 2: a Translation Between Representations answer that graphed the wrong quantity

The task. A square wire loop with sides L = 0.10 m and resistance R = 0.50 Ω moves to the right at a constant 2.0 m/s. It passes through a region 0.30 m wide containing a uniform 0.40 T magnetic field directed into the page, with no field outside the region. Its front edge reaches the field at t = 0.

• (a) Draw the loop when it is partly in the field, showing the direction of the induced current.
• (b) Derive an expression for the size of the induced current while the loop enters the field.
• (c) Sketch the induced current against time until the loop has left the field, taking counterclockwise as positive.
• (d) Justify why your graph is consistent with your expression.

A student wrote:

(a) The current is clockwise.

(b) ε = BLv, so I = BLv.

(c) The current rises steadily from zero while the loop enters. It stays at its largest value while the loop is inside the field, and drops back to zero as the loop leaves.

(d) They agree because both show a current.

The bugs.

1. (a) The direction is wrong. As the loop enters, the flux into the page through it increases, so by Lenz's law the induced current makes its own field out of the page inside the loop: counterclockwise.
2. (b) The expression leaves out R, so it is an emf in volts, not a current, and it is quoted rather than derived.
3. (c) The sketch is a graph of the flux, not of the current. The current depends on how fast the flux changes, not on how much flux there is.
4. (d) "Both show a current" is not a justification: it says nothing about why the graph has its shape.

The fix.

• (a) Counterclockwise while entering, by Lenz's law.
• (b) While the loop enters, the length x of it inside the field grows at the rate v. So the flux is Φ = BLx, and it changes by ΔΦ = BLvΔt in a time Δt. Faraday's law gives ε = ΔΦ/Δt = BLv, and the loop's resistance gives I = ε/R = BLv/R = (0.40 T)(0.10 m)(2.0 m/s)/(0.50 Ω) = 0.16 A.
• (c) Entering takes L/v = 0.050 s, so I = +0.16 A from 0 to 0.050 s. The loop is then wholly inside until its front edge reaches the far side of the field, 0.30 m from where it entered, at 0.15 s. The flux is constant, so I = 0. Leaving takes another 0.050 s, from 0.15 s to 0.20 s, with the flux decreasing, so I = -0.16 A, clockwise. After 0.20 s, I = 0. The graph is a set of flat steps: +0.16 A, zero, -0.16 A, zero.
• (d) While the loop enters, B, L, v and R are all constant, so I = BLv/R is constant: a flat segment, not a ramp. The expression came from the rate at which the loop's area inside the field changes. While the loop is wholly inside, that area does not change, so the current is zero. While it leaves, the area shrinks at the same rate, giving the same size of current in the opposite direction.

Sanity check. Energy. While the current flows, the field pushes back on the moving side of the loop, the magnetic braking of Lesson 18, with a force F = BIL = (0.40 T)(0.16 A)(0.10 m) = 0.0064 N. So an external force must supply power Fv = (0.0064 N)(2.0 m/s) = 0.013 W to keep v constant. That equals the rate of heating in the wire, I²R = (0.16 A)²(0.50 Ω) = 0.013 W. While the loop is wholly inside, there is no current, no braking force and no heating.

The variant that flips the result. Cut a small gap in one side of the loop. While it enters, the emf is still BLv = 0.080 V, but with the circuit open no current can flow. The graph of current is zero throughout, while a graph of emf would keep the stepped shape. Read the question: the current and the emf now have different answers.

Bug hunt 3: an Experimental Design and Analysis answer with no single variable

The task. Design an experiment to find the focal length of a magnifying glass, using a lamp as the object, a white card as a screen and a metre stick.

• (a) Describe your procedure: what you would vary, what you would measure, and how you would reduce uncertainty.
• (b) Describe how you would analyse the data to find the focal length.
• (c) A student took the measurements below. Choose quantities to graph that should give a straight line, plot them, draw a best-fit line, and use it to find the focal length.

These values were made up for this lesson to behave like real measurements, scatter included; they are illustrative, not data from an exam or an experiment.

A student wrote:

(a) Move the lens and the card around until the picture of the lamp is sharp, then measure the distances. Do this for different positions.

(b) Plot s_i against s_o; the focal length is where the graph crosses an axis.

(c) The points make a curve, so I drew a smooth curve, and its slope is 1/f.

The bugs.

1. (a) Nothing is controlled. Moving the lens and the card together changes two distances at once, so no single quantity is the independent variable. The CED expects a design that varies one quantity and measures its effect on one other. There is no repetition, and no statement of where the distances are measured from.
2. (b) For real images, a graph of s_i against s_o is a curve that never crosses either axis: it approaches the lines s_o = f and s_i = f without reaching them, so reading f from it means guessing where it is heading.
3. (c) A curve has no single slope, and nothing in the thin-lens equation makes any slope equal to 1/f.

The fix.

• (a) First find a rough focal length by focusing the image of a distant window onto the card; for a distant object the image distance is close to f. Then fix the lens in a holder. Set the lamp at a chosen object distance s_o, the independent variable, measured from the centre of the lens. Step it through at least six values, here from 15 cm to 50 cm. For each, move only the card until the image of the lamp is sharpest, and record s_i, the dependent variable, also from the centre of the lens. Find the sharpest position three times, approaching from nearer and from farther, and average: judging sharpness is the largest source of uncertainty.
• (b) The thin-lens equation, 1/s_o + 1/s_i = 1/f, says that a graph of 1/s_i against 1/s_o is a straight line with slope -1 and vertical intercept 1/f; its horizontal intercept is also 1/f. Choosing variables that turn a curve into a straight line in this way is called linearising the data.
• (c) The transformed values are in the table below.

Plotted, the points lie close to a straight line. The best-fit line has a slope of -1.01, close to the -1 the thin-lens equation predicts, which supports the model. Its vertical intercept is 0.1006 cm^-1, so f = 1/(0.1006 cm^-1) = 9.94 cm. Its horizontal intercept, 0.0993 cm^-1, gives 10.1 cm. The focal length is 10.0 cm to within about 1 percent.

Sanity check. Each pair should satisfy 1/s_o + 1/s_i = 1/f, and the six sums run from 0.0992 to 0.1009 cm^-1, all close to 0.100 cm^-1. A 10 cm focal length matches the magnifying glass of Lesson 21's worked problem.

The variant that flips the result. Give the same student a diverging lens. It never forms a real image, so there is no position of the card at which the image is sharp, and the method fails before any data are taken. A design has to suit the equipment, which is why design questions state exactly what is available.

Safety. Never use the Sun as the object, and never look at the Sun through a lens. A magnifying glass focuses sunlight into a spot hot enough to burn skin or start a fire. A lamp or a bright window is all the experiment needs.

Bug hunt 4: a Qualitative/Quantitative Translation answer that mistook brightness for energy

The task. Light of a single frequency f falls on a metal with work function φ in an evacuated tube, and electrons are emitted. The stopping potential V_s is measured.

• (a) The intensity of the light is doubled and its frequency kept the same. Does V_s increase, decrease or stay the same? Justify your claim in words, without equations.
• (b) Derive an expression for V_s in terms of f, φ, Planck's constant h and the elementary charge e.
• (c) Explain how your expression in (b) is consistent with your claim in (a).

A student wrote:

(a) It increases, because brighter light carries more energy, so the electrons come out faster.

(b) eV = hf - φ, so V_s = hf - φ.

(c) My answers agree, because more energy means a larger stopping voltage.

The bugs.

1. (a) The claim is wrong, and the reasoning uses the wave picture that the photoelectric effect overturned in Lesson 27. Brighter light of the same frequency means more photons each second, not more energy per photon.
2. (b) The last step divides only one side by e. The quantity hf - φ is an energy, in joules or electron volts, and cannot equal a potential in volts.
3. (c) The consistency argument is circular: it repeats the claim instead of pointing to what the equation contains.

The fix.

• (a) It stays the same. Each emitted electron gets its energy from a single photon, and a photon's energy depends only on the light's frequency. Doubling the intensity doubles the number of photons arriving each second, so twice as many electrons are emitted. But the most energetic of them have the same kinetic energy as before. The stopping potential measures that maximum kinetic energy, so it does not change.
• (b) Start from conservation of energy for the most energetic electron. Of the photon's energy hf, at least φ goes to freeing the electron, and the rest is kinetic energy, K_max = hf - φ. To stop that electron, the potential difference must remove all of it: eV_s = K_max. So V_s = (hf - φ)/e.
• (c) The expression contains f, φ, h and e, and the intensity appears nowhere in it. So the stopping potential cannot depend on the intensity, which is the claim in (a). The intensity changes the number of electrons, which shows up in the current, not in V_s.

Sanity check. Put in numbers. Light of 400 nm carries hc/λ = (1240 eV·nm)/(400 nm) = 3.10 eV per photon; for φ = 2.30 eV, V_s = (3.10 eV - 2.30 eV)/e = 0.80 V. Doubling the intensity leaves both 3.10 eV and 2.30 eV unchanged, so V_s stays at 0.80 V.

The variant that flips the result. Double the frequency instead, keeping the intensity. Each photon now carries 6.20 eV, and V_s = (6.20 eV - 2.30 eV)/e = 3.90 V. The claim flips from "stays the same" to "increases", and by a factor of nearly 5, not 2, because the work function is subtracted before anything is divided. A claim that V_s doubles would also be wrong, which the equation reveals and intuition does not.

Habits for the multiple-choice section

Forty-two questions in 85 minutes is about two minutes each, so the multiple-choice section rewards habits more than heroics.

• Read what is asked for: a magnitude or a direction, a value or a ratio, the quantity before or after the change.
• For "by what factor" questions, write the relationship and scale it rather than computing absolute values: doubling the separation of two charges quarters the force between them, because F = kq₁q₂/r².
• Use conservation laws and units as filters. An option that creates energy, leaves charge or nucleon number unbalanced, or has the wrong units can be crossed out before any calculation.
• Test the extremes. If a resistance goes to zero, or a frequency drops below threshold, which options still make sense?
• Do not let one question take five minutes. Make your best choice, mark it, and come back if time allows.

In short: every free-response part is a small contract. The verb says what to deliver. A derivation starts from a law, and a calculation carries its units. A graph has to agree with the equation beside it, and a justification is a chain of physical reasons, not a louder version of the claim.

Common misconceptions

• "A correct number is enough." A calculation earns credit for the expression, the substituted values and the units as well as the result, and a justification needs reasoning in words.
• "Derive means copying the formula from the equation tables." A derivation starts from a fundamental law or relationship and shows the steps to the result.
• "A sketch has to be drawn to scale." A sketch shows relationships and key features, such as where a graph is flat, where it changes and its sign. Plotting data is the task that needs a scale.
• "A good experiment changes as many things as possible." A sound design varies one quantity and measures its effect on one other, with repeated measurements, so that the data can answer one question.
• "Justify means stating the claim more firmly." A justification connects the claim to physics: a law, a definition, a relationship or evidence, step by step.
• "If two of my answers disagree, it is safest to leave both." Questions often ask whether parts agree. A disagreement is the signal to find which reasoning is wrong and fix it.

Summing up

The AP Physics 2 exam, from May 2027, has 42 multiple-choice questions in 85 minutes and four free-response questions in 95 minutes. Each section is half the score, and calculators and equation tables are allowed. The free-response questions are four fixed forms: Mathematical Routines (10 points), Translation Between Representations (12), Experimental Design and Analysis (10) and Qualitative/Quantitative Translation (8). Their task verbs say exactly what will earn credit.

The four debugged answers show the common failures and their fixes. Derive from a law: the loop rule gives V_T = εR₁/(R₁ + r), and an extra parallel branch lowers V_T from 8.0 V to 7.2 V. Graph the right quantity: induced current follows the rate of change of flux, giving flat steps of +0.16 A, zero and -0.16 A. Design with one independent variable and linearise: 1/s_i against 1/s_o gives slope -1 and f = 10.0 cm. Translate words and equations both ways: V_s = (hf - φ)/e contains no intensity, so brighter light leaves the stopping potential unchanged. That is the course: seven units of physics, each with a model you can calculate with and defend in words.

Sources

1. College Board. (2024). AP Physics 2: Algebra-Based course and exam description, Exam Information (exam overview, multiple-choice weightings, the four free-response question types with their points and suggested times, and the task verbs), with the 2026 clarifications. apcentral.collegeboard.org
2. College Board. (n.d.). AP Physics 2: Algebra-Based exam (the 2027 exam date; hybrid digital format; section timing; calculators and reference materials). AP Central. apcentral.collegeboard.org
3. OpenStax. (2022). 21.2 Electromotive force: Terminal voltage. In College Physics 2e. openstax.org
4. OpenStax. (2022). 23.2 Faraday's law of induction: Lenz's law; and 23.3 Motional emf. In College Physics 2e. openstax.org
5. OpenStax. (2022). 25.6 Image formation by lenses (the thin-lens equation). In College Physics 2e. openstax.org
6. OpenStax. (2022). 29.2 The photoelectric effect. In College Physics 2e. openstax.org