Ten calculus-based lessons across the six units: fields and Gauss's law, potential, conductors and capacitors, circuits, magnetism and induction, with interactive explorers.
A reasoning guide to AP Physics C: Electricity and Magnetism, not a full course. Unit titles and weightings follow the College Board course page; the exam format shown on the exam page for May 2027 is 42 multiple-choice questions in 85 minutes and 4 free-response questions in 95 minutes. All numbers in examples are invented for practice, with standard SI constants.
Calculus at the same time or before, and a mechanics-based physics course (AP Physics C: Mechanics or similar).
Course outline
Coulomb's law, fields and integrating over a charge distribution
Compute electric fields from point charges and from continuous distributions by superposition and integration.
Gauss's law and symmetry
Use Gauss's law to find fields that have spherical, cylindrical or planar symmetry.
Electric potential and potential energy
Relate potential to field, compute potentials from charges, and find energies.
Conductors in electrostatic equilibrium
Use the properties of conductors to find charge distributions and fields.
Capacitors, dielectrics and stored energy
Compute capacitance, charge and energy, and combine capacitors.
DC circuits: Ohm's law, Kirchhoff's rules and power
Analyze resistor networks with Ohm's law and the two Kirchhoff rules.
RC circuits: charging and discharging
Solve the exponential behavior of a capacitor with a resistor.
Magnetic force and the motion of charges
Find the force on moving charges and the radius and frequency of circular motion.
Magnetic fields from currents: Biot-Savart and Ampere
Compute fields of wires and solenoids using the Biot-Savart law and Ampere's law.
Electromagnetic induction and Lenz's law
Apply Faraday's law to changing flux and moving conductors.
Sources and curriculum note
Reviewed October 7, 2026. Confirm format on the College Board site for your exam year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Coulomb's law, fields and integrating over a charge distribution
Learning goal: Compute electric fields from point charges and from continuous distributions by superposition and integration.
Two point charges exert a force of magnitude F = k q1 q2 / r^2 along the line joining them, with k = 8.99 x 10^9 N m^2 / C^2, equal to 1 / (4 pi epsilon0). Like charges repel and unlike charges attract. The electric field of a point charge is E = k q / r^2, directed away from positive charge and toward negative charge, and the force on a test charge is F = q E.
Fields add as vectors. For several charges, find each field vector at the point, break it into components, and add the components. Draw the diagram first and decide the sign of each component before you calculate. Magnitudes alone will not give the direction.
For a continuous distribution, split the charge into small pieces dq, write the field dE of each piece, and integrate. Use the symmetry to cancel components. For a ring of charge Q and radius a, the field on the axis at distance x is E = k Q x / (x^2 + a^2)^(3/2). The perpendicular parts cancel in pairs, so only the axial part is left, and it vanishes at the center and far away it behaves like a point charge.
Check limits as a habit. At x much larger than a the ring formula tends to k Q / x^2. At x = 0 it gives zero, which fits symmetry. If your integral fails these two checks, the setup is wrong.
Worked example
Two charges +2.0 nC and +2.0 nC sit 0.10 m apart. Force between them in newtons?
k q1 q2 / r^2.
8.99e9 x 2e-9 x 2e-9.
Divide by 0.01.
3.6 x 10^-6 N, repulsive.
Practice problem and solution
Two +2.0 nC charges are 0.10 m apart. Force in micronewtons, to 1 decimal place? Use k = 8.99e9.
F = 8.99e9 x 4e-18 / 0.01 = 3.6e-6 N.
Mental model: E = kq/r^2. Add fields as vectors. For a ring on the axis, E = kQx/(x^2+a^2)^(3/2).
Common trap: Adding magnitudes of vectors that point in different directions.
2. Gauss's law and symmetry
Learning goal: Use Gauss's law to find fields that have spherical, cylindrical or planar symmetry.
Electric flux is the surface integral of E dot dA. Gauss's law says the net flux through any closed surface equals the enclosed charge divided by epsilon0. It is always true, but it gives the field only when symmetry lets you pull E out of the integral.
Pick a Gaussian surface where E is constant in magnitude over the part that matters and either parallel or perpendicular to the surface elsewhere. For a sphere, use a concentric sphere: E x 4 pi r^2 = Q_enclosed / epsilon0, so E = k Q_enclosed / r^2. For an infinite line charge with density lambda, use a coaxial cylinder: E = lambda / (2 pi epsilon0 r). For an infinite sheet with density sigma, E = sigma / (2 epsilon0) on each side.
For a uniformly charged solid sphere of total charge Q and radius R, the field outside is k Q / r^2. Inside, only the charge within radius r counts, which is Q r^3 / R^3, so E = k Q r / R^3. The field grows linearly with r to a peak at the surface, then falls as 1 / r^2.
A common error is to use Gauss's law when there is no symmetry, or to include charge outside the surface. Only the enclosed charge counts, but the field at each point comes from all charges.
Worked example
A solid sphere has Q = 8.0 nC and R = 4.0 cm. Field at r = 2.0 cm?
Inside: E = k Q r / R^3.
Q r / R^3 = 8e-9 x 0.02 / 6.4e-5.
k times that.
Check with Q r^3 / R^3 enclosed.
Practice problem and solution
Q = 8.0 nC, R = 4.0 cm, r = 2.0 cm. Field in N/C, as a whole number?
E = k Q r / R^3.
Mental model: Enclosed charge sets the flux; symmetry gives E. Inside a solid sphere E rises linearly.
Common trap: Using Gauss's law without symmetry.
3. Electric potential and potential energy
Learning goal: Relate potential to field, compute potentials from charges, and find energies.
Electric potential V is the potential energy per unit charge. For a point charge, V = k q / r with zero at infinity. Potentials are scalars, so they add as plain numbers with their signs. That makes potential easier than field for several charges.
The field is minus the gradient of the potential. In one dimension E = -dV/dx, so the field points toward lower potential and is large where V changes quickly. Going the other way, the potential difference between two points is minus the integral of E dot dl. Work done by the field on a charge q moving between points is q times the drop in potential.
The energy of a system of point charges is the sum over pairs, U = sum of k q_i q_j / r_ij. For two charges this is k q1 q2 / r. A positive result means work was needed to bring them together from far apart. A proton in a potential difference of 1 volt gains 1 electron volt, about 1.6 x 10^-19 J.
Equipotential surfaces are perpendicular to field lines. Moving a charge along an equipotential takes no work. Use that to sketch fields from potential maps and the reverse.
Worked example
Charges +3.0 nC at x = 0 and -3.0 nC at x = 0.20 m. Potential at the midpoint?
Each is 0.10 m away.
k(3e-9)/0.1 = +270 V.
Second is -270 V.
Sum is 0 V.
Practice problem and solution
A +3.0 nC and a +5.0 nC charge are 0.30 m apart. Potential energy in microjoules, to 2 decimals? Use k = 8.99e9.
U = k q1 q2 / r = 8.99e9 x 15e-18 / 0.30 = 4.49e-7 J.
Mental model: V = kq/r adds as a scalar. E = -dV/dx. U = k q1 q2 / r for each pair.
Common trap: Adding potentials like vectors.
4. Conductors in electrostatic equilibrium
Learning goal: Use the properties of conductors to find charge distributions and fields.
In electrostatic equilibrium, the field inside a conductor is zero, so the whole conductor is at one potential and any excess charge sits on the surface. Just outside a surface the field is perpendicular to it with magnitude sigma / epsilon0, twice the value for an isolated sheet because the conductor's other side contributes no field.
A cavity in a conductor can shield its inside from outside fields. If a charge q is placed inside a cavity, the cavity wall gets induced charge -q so that the field in the conductor is zero. By conservation, the outer surface carries the conductor's net charge plus q.
For a charged point q at the center of a spherical shell with inner radius a, outer radius b and net charge Q_s, Gauss's law gives -q on the inner surface and q + Q_s on the outer surface. Outside the shell the field is k (q + Q_s) / r^2. Inside the conducting material the field is zero.
For two connected conductors, charge flows until their potentials match. Two spheres joined by a thin wire share charge so that k Q1 / R1 = k Q2 / R2, so the smaller sphere gets the larger surface charge density.
Worked example
A +4.0 nC charge is at the center of a neutral spherical shell. Charge on the outer surface?
Inner surface gets -4.0 nC.
Net charge is zero.
Outer = 0 - (-4.0).
+4.0 nC.
Practice problem and solution
A +4.0 nC point charge is at the center of a shell with net charge -1.0 nC. What is the charge on the outer surface in nC?
Outer = q + Q_s = 4.0 - 1.0 = 3.0 nC.
Mental model: Field inside a conductor is zero. Inner surface = -q, outer = q + Q_shell.
Common trap: Forgetting the induced charge on the inner surface.
5. Capacitors, dielectrics and stored energy
Learning goal: Compute capacitance, charge and energy, and combine capacitors.
Capacitance is the charge stored per volt, C = Q / V. For parallel plates of area A and separation d with vacuum between them, C = epsilon0 A / d. Capacitance depends only on geometry and the material, not on the charge or voltage.
A dielectric of constant kappa between the plates multiplies the capacitance by kappa. If the battery stays connected, the voltage is fixed, so charge increases by kappa. If the battery is disconnected, the charge is fixed, so voltage drops by kappa. Keep the two cases apart.
The energy stored is U = (1/2) C V^2 = Q^2 / (2C) = (1/2) Q V. In a parallel-plate capacitor the energy density is (1/2) epsilon0 E^2, which is the energy per unit volume in the field. Capacitors in parallel add, C = C1 + C2, and in series add as reciprocals, 1 / C = 1 / C1 + 1 / C2.
Use units carefully: area in square meters, separation in meters. A 1 cm squared plate at 1 mm separation has C = epsilon0 x 1e-4 / 1e-3 = 0.885 pF.
Worked example
A parallel-plate capacitor has A = 100 cm^2 and d = 1.0 mm, vacuum. Capacitance in pF?
A = 0.01 m^2.
d = 0.001 m.
eps0 A / d.
88.5 pF.
Practice problem and solution
A 20 pF capacitor is charged to 50 V. Energy in nanojoules?
U = 0.5 x 20e-12 x 2500 = 2.5e-8 J.
Mental model: C = eps0 kappa A / d. U = (1/2) C V^2. Parallel adds, series adds reciprocals.
Common trap: Mixing up the connected and isolated dielectric cases.
6. DC circuits: Ohm's law, Kirchhoff's rules and power
Learning goal: Analyze resistor networks with Ohm's law and the two Kirchhoff rules.
Ohm's law says V = I R for an ohmic resistor. Current is the rate of flow of charge, I = dQ / dt, and resistance depends on resistivity and shape, R = rho L / A. Real batteries have internal resistance, so terminal voltage is the emf minus I r.
Resistors in series carry the same current and add: R = R1 + R2. Resistors in parallel share the same voltage and add as reciprocals: 1 / R = 1 / R1 + 1 / R2. Reduce the circuit step by step, then work back to find each current and voltage.
Kirchhoff's junction rule says the currents into a junction sum to zero, which is conservation of charge. The loop rule says the voltage changes around any closed loop sum to zero, which is conservation of energy. Choose current directions, write one equation per loop and junction, and solve. A negative answer only means the real current is opposite to your guess.
Power dissipated in a resistor is P = I V = I^2 R = V^2 / R. A 12 V battery across 4 ohms and 6 ohms in series drives 1.2 A, with 5.76 W in the 4 ohm and 8.64 W in the 6 ohm resistor. Always check that power supplied equals power dissipated.
Worked example
12 V across 4 ohm and 6 ohm in series. Current?
Series: R = 10 ohm.
I = V / R.
12 / 10.
1.2 A.
Practice problem and solution
Resistors of 6 ohm and 3 ohm in parallel across 12 V. Total current in A?
R = 18/9 = 2 ohm, I = 12 / 2 = 6 A.
Mental model: Series: R adds, same I. Parallel: 1/R adds, same V. Kirchhoff: sums of I at junctions, V around loops.
Common trap: Using the wrong current direction convention inconsistently.
7. RC circuits: charging and discharging
Learning goal: Solve the exponential behavior of a capacitor with a resistor.
A resistor in series with a capacitor makes the charge change exponentially. Kirchhoff's loop rule gives emf = I R + Q / C with I = dQ / dt. The solution has the time constant tau = R C, the time for the change to reach about 63 percent of its way.
When charging through R from a battery of emf E, Q(t) = C E (1 - e^(-t / tau)) and I(t) = (E / R) e^(-t / tau). The current is largest at the start, E / R, because the capacitor acts as a short, and it decays to zero as the capacitor acts as an open circuit.
When discharging, Q(t) = Q0 e^(-t / tau) and the current has the same exponential form. After one time constant the quantity is 37 percent of its start, after two it is 14 percent, and after five it is under 1 percent. In many practical cases five time constants is treated as essentially finished.
Unit check: ohms times farads is seconds. A 10 kilohm resistor with 100 microfarads has tau = 1.0 s. To verify an RC result, check t = 0 and t much larger than tau, then check the half-way value at t = 0.69 tau.
Worked example
R = 10 kohm and C = 100 uF. Time constant?
R = 1e4 ohm.
C = 1e-4 F.
Product.
1.0 s.
Practice problem and solution
A capacitor discharges with tau = 2.0 s. Fraction of the initial charge left after 2.0 s, as a decimal to 2 places?
e^(-1) = 0.368.
Mental model: tau = RC. Charging gives 63 percent per tau, discharging leaves 37 percent per tau.
Common trap: Treating the current as constant during charging.
8. Magnetic force and the motion of charges
Learning goal: Find the force on moving charges and the radius and frequency of circular motion.
The magnetic force on a charge q moving with velocity v in a field B is F = q v x B. Its magnitude is q v B sin(theta), and its direction is perpendicular to both v and B. Because it is perpendicular to the velocity, a magnetic force does no work and cannot change the speed, only the direction.
For a positive charge use the right hand: fingers along v, curl toward B, and the thumb gives the force. For a negative charge the force is reversed. Pay attention to the direction of B into or out of the page.
If v is perpendicular to B, the charge moves in a circle. Setting q v B = m v^2 / r gives r = m v / (q B). The period T = 2 pi m / (q B) does not depend on speed, so the cyclotron frequency f = q B / (2 pi m) is the same for any speed in a given field, as long as speeds stay small compared with that of light.
For a proton in a 1.0 T field moving at 1.0 x 10^6 m/s, r = (1.67e-27 x 1e6) / (1.60e-19 x 1.0) = 1.0 cm and f is about 15 MHz. An electron with the same speed has a radius about 1,836 times smaller. A current-carrying wire in a field feels F = I L x B, which is the same force summed over many charges.
Worked example
A proton moves at 2.0 x 10^6 m/s perpendicular to a 0.50 T field. Radius in cm?
r = m v / (q B).
1.67e-27 x 2e6 / (1.60e-19 x 0.5).
= 0.0418 m.
4.2 cm.
Practice problem and solution
A proton at 2.0e6 m/s moves perpendicular to 0.50 T. Radius in cm to 1 decimal?
r = m v / (q B) = 0.0418 m.
Mental model: F = qvB sin theta, no work. r = mv/(qB), f = qB/(2 pi m).
Common trap: Thinking the speed changes in a magnetic field.
9. Magnetic fields from currents: Biot-Savart and Ampere
Learning goal: Compute fields of wires and solenoids using the Biot-Savart law and Ampere's law.
The Biot-Savart law gives the field from a small current element: dB = (mu0 / 4 pi) I dl x r_hat / r^2, with mu0 = 4 pi x 10^-7 T m / A. You add up all elements by integrating. It always works but may need an integral.
For a long straight wire, the result is B = mu0 I / (2 pi r), with field lines forming circles around the wire in the direction given by the right-hand rule: thumb along the current, fingers curl along B. At 10 A and 10 cm the field is 2e-7 x 10 / 0.1 = 20 microtesla.
Ampere's law says the line integral of B dot dl around a closed loop equals mu0 times the enclosed current. Like Gauss's law it gives the field directly only when symmetry makes B constant along the path. For a long solenoid with n turns per meter, the field inside is B = mu0 n I, nearly uniform, and the field outside is close to zero.
Use Ampere's law for the wire, the solenoid and a thick cable, and Biot-Savart for the loop. For a circular loop of radius a the field at its center is mu0 I / (2 a). Parallel wires with currents in the same direction attract, which is how the ampere was once defined.
Worked example
Long wire carrying 10 A. Field 10 cm away in microtesla?
B = mu0 I / (2 pi r).
2e-7 x 10 / 0.10.
= 2e-5 T.
20 microtesla.
Practice problem and solution
A solenoid has 800 turns per meter and carries 2.0 A. Field inside in mT, to 2 decimals?
B = mu0 n I = 1.2566e-6 x 800 x 2.0 = 2.01 mT.
Mental model: Wire: mu0 I/(2 pi r). Solenoid: mu0 n I. Ampere's law needs symmetry.
Common trap: Using Ampere's law without symmetry.
10. Electromagnetic induction and Lenz's law
Learning goal: Apply Faraday's law to changing flux and moving conductors.
Magnetic flux through a surface is the integral of B dot dA. Faraday's law says the induced emf in a loop is minus the rate of change of flux through it, emf = -d(Phi)/dt. Flux can change if B changes, if the area changes or if the angle changes.
Lenz's law gives the sign: the induced current opposes the change in flux. If flux into the page increases, the induced current makes a field out of the page inside the loop. The minus sign in Faraday's law is that statement, and it guards against violating energy conservation.
A straight rod of length L moving at speed v perpendicular to a field B has motional emf = B L v. If it slides on rails connected to a resistor R, the current is B L v / R and the magnetic force on the rod is B^2 L^2 v / R, opposing the motion. The power supplied by whoever pulls it equals the power dissipated, B^2 L^2 v^2 / R.
Inductors store energy in their field, U = (1/2) L I^2, and in an RL circuit the current grows with time constant L / R. Use units: tesla times meter squared per second is volts. In every problem, state whether flux is increasing or decreasing before writing the direction.
Worked example
A 0.50 m rod moves at 4.0 m/s through a 0.20 T field. Motional emf?
emf = B L v.
0.20 x 0.50 x 4.0.
= 0.40 V.
Across the rod.
Practice problem and solution
A 0.50 m rod on rails moves at 4.0 m/s in 0.20 T with a 2.0 ohm resistor. Current in mA?
I = B L v / R = 0.40 / 2.0 = 0.20 A.
Mental model: emf = -dPhi/dt. Rod on rails: emf = BLv, force = B^2 L^2 v / R, opposing motion.