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AP Physics C: E&M mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 30 questions and 4 free-response questions · about 140 minutes

A full-length practice test built from our exam-style questions on every unit of AP Physics C: E&M. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page
  1. Before you start
  2. Section I
  3. Section II
  4. After the mock

Before you start

Section I: 30 exam-style questions

Q1Exam-style
A closed surface surrounds a +5.0 nC charge. What is the electric flux through the surface?
N·m²/C

Hint

Gauss's law: the flux depends only on the charge enclosed.

Worked solution
  1. Φ = qenc/ε0.
  2. = 5.0 × 10−9/8.85 × 10−12.
  3. Φ = 565 N·m²/C, whatever the shape of the surface.
Q2Exam-style
A long straight wire carries a uniform charge of 2.0 μC per metre. What is the field 0.10 m from the wire?
N/C

Hint

Use a cylindrical Gaussian surface around the wire.

Worked solution
  1. Gauss's law with a cylinder of radius r and length L: E(2πrL) = λL/ε0.
  2. E = λ/(2πε0r) = 2.0 × 10−6/(2π × 8.85 × 10−12 × 0.10).
  3. E = 3.60 × 105 N/C.
Q3Exam-style
A thin spherical shell carries charge spread uniformly over its surface. The electric field inside the shell is:

Hint

Draw a Gaussian sphere inside the shell. How much charge does it enclose?

Worked solution
  1. A Gaussian sphere inside the shell encloses no charge.
  2. By symmetry E is the same all over that sphere, so E × 4πr2 = 0.
  3. So E = 0 everywhere inside: A.
Q4Challenge
A solid sphere of radius 0.20 m has a charge of 8.0 nC spread uniformly through its volume. What is the field 0.10 m from the centre?
N/C

Hint

Inside a uniform sphere, only the charge within radius r counts, and it grows as r3.

Worked solution
  1. Enclosed charge: Q(r/R)3. So E = kQr/R3.
  2. E = 9.0 × 109 × 8.0 × 10−9 × 0.10/0.203 = 7.2/0.008.
  3. E = 900 N/C.
Q5Exam-style
A large flat sheet carries 1.77 × 10−9 C/m². What is the field near it, away from the edges?
N/C

Hint

Use a Gaussian “pillbox” through the sheet.

Worked solution
  1. The flux leaves through both faces: E × 2A = σA/ε0.
  2. E = σ/(2ε0) = 1.77 × 10−9/(1.77 × 10−11).
  3. E = 100 N/C, independent of distance.
Q6Exam-style
The potential along the x-axis is V(x) = 4x2 volts (x in metres). Find Ex at x = 2 m.
V/m

Hint

Ex = −dV/dx.

Worked solution
  1. dV/dx = 8x.
  2. Ex = −8 × 2 = −16 V/m: the field points in the negative x direction.
Q7Exam-style
An electron starts from rest and is accelerated through 100 V. What is its speed?
m/s

Hint

The energy gained is eΔV, and it becomes kinetic energy.

Worked solution
  1. eΔV = ½mv2, so v = √(2eΔV/m).
  2. v = √(2 × 1.6 × 10−19 × 100/9.11 × 10−31).
  3. v = 5.93 × 106 m/s.
Q8Exam-style
What is the electric potential energy of a +3.0 μC charge and a −2.0 μC charge 0.20 m apart?
J

Hint

U = kq1q2/r, keeping the signs.

Worked solution
  1. U = 9.0 × 109 × (3.0 × 10−6)(−2.0 × 10−6)/0.20.
  2. U = −0.27 J. Negative: the pair is bound.
Q9Exam-style
Four +1.0 nC charges sit at the corners of a square of side 0.20 m. What is the potential at the centre?
V

Hint

Potential is a scalar: add the four contributions.

Worked solution
  1. Each charge is half a diagonal away: 0.20/√2 = 0.1414 m.
  2. Each gives kq/r = 9.0/0.1414 = 63.6 V.
  3. Total = 4 × 63.6 = 255 V (while the field at the centre is zero).
Q10Exam-style
In a uniform field of 300 N/C, what is the potential difference between two points 0.050 m apart along the field?
V

Hint

In a uniform field, |ΔV| = Ed.

Worked solution
  1. |ΔV| = 300 × 0.050.
  2. = 15 V, lower in the direction of the field.
Q11Exam-style
A parallel-plate capacitor has plates of area 0.010 m², 1.0 mm apart, in air. What is its capacitance, in picofarads?
pF

Hint

C = ε0A/d, then convert to pF.

Worked solution
  1. C = 8.85 × 10−12 × 0.010/1.0 × 10−3.
  2. C = 8.85 × 10−11 F.
  3. = 88.5 pF.
Q12Exam-style
A 3.0 μF and a 6.0 μF capacitor are connected in series. What is their combined capacitance, in μF?
μF

Hint

In series, the reciprocals add.

Worked solution
  1. 1/C = 1/3 + 1/6 = 1/2.
  2. C = 2.0 μF, smaller than either one.
Q13Exam-style
The same two capacitors (3.0 μF and 6.0 μF) are connected in parallel instead. What is their combined capacitance, in μF?
μF

Hint

In parallel, capacitances simply add.

Worked solution
  1. C = 3.0 + 6.0 = 9.0 μF.
Q14Exam-style
A capacitor stays connected to a battery while a dielectric (κ > 1) is slid between its plates. What happens?

Hint

The battery fixes the voltage.

Worked solution
  1. The dielectric multiplies C by κ.
  2. The battery holds V constant, so Q = CV increases.
  3. The answer is A. (Disconnected, Q would stay fixed and V would fall.)
Q15Challenge
What is the capacitance of an isolated conducting sphere of radius 0.10 m, in picofarads?
pF

Hint

The sphere's potential is kQ/R, so C = Q/V.

Worked solution
  1. V = kQ/R, so C = R/k = 4πε0R.
  2. C = 4π × 8.85 × 10−12 × 0.10 = 1.11 × 10−11 F.
  3. = 11.1 pF.
Q16Exam-style
A current of 2.0 A flows through a 6.0 Ω resistor. What power does it dissipate?
W

Hint

P = I2R.

Worked solution
  1. P = 2.02 × 6.0 = 24 W.
Q17Exam-style
What is the time constant of a circuit with R = 2.0 kΩ and C = 50 μF?
s

Hint

τ = RC, in base units.

Worked solution
  1. τ = 2000 × 50 × 10−6.
  2. τ = 0.10 s.
Q18Exam-style
A capacitor charges through a resistor from a 12 V battery, starting uncharged. What is the capacitor's voltage after one time constant?
V

Hint

After one time constant, the capacitor has reached 1 − e−1 of its final value.

Worked solution
  1. VC = 12(1 − e−1) = 12 × 0.632.
  2. VC = 7.59 V.
Q19Exam-style
A 12 V battery with internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. What is the terminal voltage of the battery?
V

Hint

Find the current first, using the total resistance.

Worked solution
  1. I = 12/(5.5 + 0.50) = 2.0 A.
  2. Terminal voltage = ε − Ir = 12 − 2.0 × 0.50.
  3. = 11 V.
Q20Challenge
A 100 μF capacitor charged to 9.0 V discharges through a 10 kΩ resistor. How long until its voltage falls to 4.5 V?
s

Hint

Voltage falls as e−t/RC. Half the voltage means one “half-life”.

Worked solution
  1. RC = 104 × 10−4 = 1.0 s.
  2. 4.5 = 9.0e−t/RC, so t = RC ln 2.
  3. t = 0.693 s.
Q21Exam-style
An electron moves at 1.0 × 106 m/s perpendicular to a 0.010 T field. What is the radius of its circular path?
m

Hint

The magnetic force provides the centripetal force: qvB = mv2/r.

Worked solution
  1. r = mv/(qB).
  2. r = 9.11 × 10−31 × 1.0 × 106/(1.6 × 10−19 × 0.010).
  3. r = 5.69 × 10−4 m.
Q22Exam-style
Two long parallel wires 0.10 m apart each carry 5.0 A. What is the force per metre between them?
N/m

Hint

One wire sits in the other's field: F/L = I2B1.

Worked solution
  1. F/L = μ0I1I2/(2πd).
  2. = 4π × 10−7 × 25/(2π × 0.10).
  3. F/L = 5.0 × 10−5 N/m, attractive if the currents are in the same direction.
Q23Exam-style
A long solenoid has 1000 turns per metre and carries 2.0 A. What is the field inside it?
T

Hint

B = μ0nI.

Worked solution
  1. B = 4π × 10−7 × 1000 × 2.0.
  2. B = 2.51 × 10−3 T.
Q24Exam-style
A 0.20 m wire carrying 3.0 A sits at right angles to a 0.40 T field. What is the force on it?
N

Hint

F = ILB.

Worked solution
  1. F = 3.0 × 0.20 × 0.40 = 0.24 N.
Q25Challenge
A long, thick wire of radius R carries current spread uniformly across its cross-section. Inside the wire (r < R), the field:

Hint

Apply Ampère's law to a circle of radius r inside the wire.

Worked solution
  1. The enclosed current is I(r/R)2.
  2. B(2πr) = μ0I r2/R2, so B = μ0Ir/(2πR2).
  3. B grows in proportion to r: C.
Q26Exam-style
A 100-turn coil has its flux change by 0.020 Wb in 0.10 s. What is the average induced emf?
V

Hint

Faraday's law, with N turns.

Worked solution
  1. |ε| = NΔΦ/Δt = 100 × 0.020/0.10.
  2. = 20 V.
Q27Exam-style
The current in a 0.50 H inductor changes at 4.0 A/s. What emf is induced?
V

Hint

ε = L dI/dt (magnitude).

Worked solution
  1. |ε| = 0.50 × 4.0 = 2.0 V.
Q28Exam-style
How much energy is stored in a 0.20 H inductor carrying 3.0 A?
J

Hint

U = ½LI2.

Worked solution
  1. U = ½ × 0.20 × 3.02 = 0.90 J.
Q29Exam-style
What is the time constant of an RL circuit with L = 0.30 H and R = 6.0 Ω?
s

Hint

τ = L/R.

Worked solution
  1. τ = 0.30/6.0 = 0.050 s.
Q30Challenge
A 12 V battery, a 4.0 Ω resistor and an inductor are connected in series, and the switch is closed at t = 0. What is the current after one time constant?
A

Hint

The current rises towards ε/R as 1 − e−t/τ.

Worked solution
  1. Final current: 12/4.0 = 3.0 A.
  2. I(τ) = 3.0(1 − e−1) = 3.0 × 0.632.
  3. I = 1.90 A.

Section II: 4 free-response questions

Free-response question 1: Electric charges, fields and Gauss's law

Mathematical routines style · 6 points · about 20 minutes

A solid insulating sphere of radius R has total charge Q spread uniformly through its volume.

  1. (a) Use Gauss's law to derive the electric field at a distance r < R from the centre.
  2. (b) Derive the field for r > R.
  3. (c) Sketch or describe the graph of E against r from r = 0 to r = 3R.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Enclosed charge Q(r/R)3. E(4πr2) = Qr3/(ε0R3), so E = Qr/(4πε0R3).

Part (b)

All the charge is enclosed: E = Q/(4πε0r2).

Part (c)

E rises linearly from 0 at the centre to a maximum at r = R, then falls as 1/r2.

Your mark: 0 of 6

Free-response question 2: Conductors and capacitors

Qualitative/quantitative translation style · 6 points · about 20 minutes

A parallel-plate capacitor with plate area A and separation d is charged by a battery of voltage V0, and then disconnected. The plates are then pulled apart to separation 2d.

  1. (a) Derive the capacitance of the original capacitor using Gauss's law.
  2. (b) After the plates are pulled apart, state whether Q, C, V and the stored energy increase, decrease or stay the same. Justify each.
  3. (c) Where does the extra stored energy come from?

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Field between the plates E = σ/ε0 = Q/(ε0A). V = Ed = Qd/(ε0A), so C = Q/V = ε0A/d.

Part (b)

Q stays the same (disconnected). C halves (C ∝ 1/d). V = Q/C doubles. U = Q2/(2C) doubles.

Part (c)

From the work done by whoever pulls the plates apart against their mutual attraction.

Your mark: 0 of 6

Free-response question 3: Electric circuits

Mathematical routines style · 7 points · about 20 minutes

An uncharged capacitor C is connected in series with a resistor R, a battery of emf ε and a switch. The switch is closed at t = 0.

  1. (a) Use Kirchhoff's loop rule to write a differential equation for the charge q on the capacitor.
  2. (b) Solve it to find q(t).
  3. (c) Find the current immediately after the switch closes, and after a long time.
  4. (d) Show that the energy dissipated in the resistor while charging equals the energy finally stored in the capacitor.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

ε − R(dq/dt) − q/C = 0.

Part (b)

Separating: dq/(Cε − q) = dt/(RC), giving q(t) = Cε(1 − e−t/RC).

Part (c)

At t = 0 the capacitor acts like a wire: I = ε/R. After a long time it is fully charged: I = 0.

Part (d)

The battery supplies Qε = Cε2; the capacitor stores ½Cε2; so the resistor dissipates the other ½Cε2.

Your mark: 0 of 7

Free-response question 4: Electromagnetic induction

Qualitative/quantitative translation style · 7 points · about 20 minutes

A conducting rod of length L slides at constant speed v along two parallel rails joined by a resistor R. A uniform field B points into the page, at right angles to the rails.

  1. (a) Derive the emf induced in the circuit.
  2. (b) Find the current, and state its direction using Lenz's law.
  3. (c) Find the external force needed to keep the rod moving at constant speed.
  4. (d) Show that the mechanical power supplied equals the electrical power dissipated in R.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

The area grows at Lv per second, so dΦ/dt = BLv and ε = BLv.

Part (b)

I = BLv/R. The flux into the page is increasing, so the induced current is anticlockwise, making a field out of the page.

Part (c)

The magnetic force on the rod is ILB = B2L2v/R, opposing the motion, so the pulling force must equal it.

Part (d)

Fv = B2L2v2/R, and I2R = (BLv/R)2R = B2L2v2/R. They are equal: energy is conserved.

Your mark: 0 of 7

After the mock

Note every question you missed and its unit, then go back to that unit's practice page: all AP Physics C: E&M units. Then try the real thing with official AP Physics C: E&M past papers.