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Electric potential worksheet with answers for AP Physics C (free PDF download)

Quick answer: Electric potential is potential energy per unit charge, V = U/q, measured in volts. This AP Physics C worksheet on electric potential gives 20 problems from easy to hard: units of potential, point charge potential and uniform field gap. The worked answers are on a separate page, and the worksheet is a free printable PDF.

Electric potential worksheet with answers for AP Physics C (free PDF download)

These twenty problems train you to move between field, potential and energy using calculus: integrating E to get V, differentiating V to get E, and using conductors and equipotentials. They progress from one-step recall to multi-step derivations in the style of AP Physics C: Electricity and Magnetism.
The set suits students on the AP course, first-year university physics and A Level learners ready for calculus. Work with k = 8.99 × 10⁹ N·m²/C².

Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 9

Problem 2 Point charge potential ★★★

1. Calculate the potential 0.30 m from a +5.0 nC point charge, taking V = 0 at infinity.
2. State the potential at 0.60 m from the same charge.

Problem 3 Uniform field gap ★★★

Aarav measures a uniform field of 1500 N/C between two parallel plates 0.020 m apart.
1. Calculate the potential difference between the plates.
2. State which plate is at the higher potential if the field points from plate A to plate B.

Problem 4 Electron accelerated ★★★

Chiamaka’s school demonstration tube accelerates electrons from rest through 250 V.
1. Calculate the kinetic energy gained, in joules.
2. Calculate the final speed (m = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C).

Problem 5 Inside a conductor ★★★

1. State the electric field inside a solid conductor in electrostatic equilibrium.
2. Explain why the whole conductor is at one potential.

Problem 6 Reading a V–x graph ★★★

The graph shows the potential along the x-axis in a region of space.
Graph of potential against position x: V falls in a straight line from 12 volts at x equal to zero to zero volts at x equal to 4.0 centimetresx (cm)V (volts)124.00
Potential along x.
1. Calculate the magnitude of E in the region.
2. State the direction of E.
3. Calculate the work done by the field on a +2.0 μC charge moved from x = 0 to x = 4.0 cm.

Problem 7 Two charges, one point ★★★

Charges of +3.0 nC and −1.0 nC are 0.40 m apart.
1. Calculate the potential at the midpoint.
2. Explain why you can add the two potentials without using vectors.

Problem 8 Field given by x² ★★★

Along the x-axis Ex = (600 V/m³)x², and V(0) = 0.
1. Integrate to find V(x).
2. Calculate V at x = 0.30 m.

Problem 9 Field from potential ★★★

A potential varies as V(x) = 40x − 150x² (V, with x in metres).
1. Derive Ex(x).
2. Calculate Ex at x = 0.10 m.
3. Find where Ex = 0.

Problem 10 Ring on its axis ★★★

A ring of radius 0.12 m carries a uniform charge of 4.0 nC.
1. Calculate the potential on the axis 0.16 m from the centre.
2. Calculate Ex at that point.

Problem 11 Assembling two charges ★★★

Wanjiru brings a 2.0 μC charge from infinity to 0.50 m from a fixed 3.0 μC charge.
1. Calculate the work she must do.
2. Explain the sign of your answer.

Problem 12 Charged sphere ★★★

A conducting sphere of radius 0.15 m is held at 3000 V.
1. Calculate its charge.
2. Calculate the field at its surface.
3. Calculate the potential 0.45 m from the centre.
4. State V and E inside.

Problem 13 Joined spheres ★★★

The two spheres shown are joined by a thin wire and share a total charge of 80 nC.
Two conducting spheres joined by a wire: a small sphere of radius 0.10 metres and a large sphere of radius 0.30 metresR₁ = 0.10 mR₂ = 0.30 mwire
Two spheres joined by a wire.
1. Explain why they reach the same potential.
2. Calculate the charge on each sphere.
3. Calculate the common potential.
4. Compare the surface fields E₁ and E₂.

Problem 14 Spacing of equipotentials ★★★

The diagram shows three equipotential lines in a flat region.
Three vertical equipotential lines labelled 20 volts, 40 volts and 60 volts; the gap between 20 and 40 volts is 2.0 centimetres and the gap between 40 and 60 volts is 1.0 centimetre20 V40 V60 V2.0 cm1.0 cm
Equipotentials at 20 V, 40 V and 60 V.
1. Calculate the average field between 20 V and 40 V.
2. Calculate the average field between 40 V and 60 V.
3. Explain where the field is stronger.
4. State the direction of the field.

Problem 15 Potential near a long wire ★★★

A long straight wire carries a uniform charge of 3.0 nC per metre, so E = 2kλ/r.
1. Derive the potential difference V(0.10 m) − V(0.40 m).
2. Calculate it and state which point is at higher potential.

Problem 16 Charged rod on its axis ★★★

A thin rod of length 0.50 m carries 6.0 nC uniformly. Point P lies on the rod’s line, 0.10 m beyond one end.
1. Show that V = k(Q/L) ln[(L + d)/d] for distance d from the near end.
2. Calculate V at P.

Problem 17 Charged disc ★★★

A disc of radius 0.20 m carries uniform surface charge 5.0 nC/m². Its potential on the axis is V = (σ/2ε₀)(√(x² + R²) − x), ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
1. Calculate V at x = 0.15 m.
2. Differentiate to find Ex and calculate it at x = 0.15 m.
3. Show that V tends to kQ/x for large x.

Problem 18 Uniform charged ball ★★★

Kofi models a cloud as a uniformly charged ball of radius 0.050 m holding 10 nC, with E = kQr/R³ inside.
1. Integrate to find V(r) inside, matching kQ/R at the surface.
2. Calculate V at the centre.
3. Calculate V at r = R/2.

Problem 19 Particle released near a charge ★★★

A 5.0 g bead carrying +2.0 μC is released from rest 0.10 m from a fixed +4.0 μC charge, on a frictionless insulating track.
1. Calculate its speed when it reaches r = 0.50 m.
2. Calculate its speed as r → ∞.
3. Explain why the speed never exceeds the second value.

Problem 20 Spot the error ★★★

Mei says: ‘Inside a hollow charged conducting shell E = 0, so V = 0 there too.’ The shell has radius 0.25 m and charge −30 nC.
1. Identify and correct her mistake.
2. Calculate the potential inside the shell.
3. Calculate the potential 0.50 m from the centre.

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