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Electric potential practice problems AP Physics C: E&M, Unit 9

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

Electric potential and potential energy, potential of point charges and charge distributions, and the link E = −dV/dx. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for electric potential

V = kq/rpotential of a point charge
U = kq1q2/rpotential energy of two charges
W = qΔVwork to move a charge
Ex = −dV/dxfield from potential
ΔV = −∫ E·drpotential from field

Constants as on the AP reference sheet: k = 9.0 × 109 N·m²/C², ε0 = 8.85 × 10−12 C²/(N·m²), μ0 = 4π × 10−7 T·m/A, e = 1.6 × 10−19 C. You can type answers like 3.6e5 or 3.6x10^5. The full sheet is in our AP Physics C equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
What is the electric potential 0.30 m from a 6.0 nC point charge?
V

Hint

V = kq/r: no square, and it is a scalar.

Worked solution
  1. V = 9.0 × 109 × 6.0 × 10−9/0.30.
  2. V = 180 V.
Q2Warm-up
How much work is needed to move a 2.0 μC charge through a potential difference of 50 V?
J

Hint

W = qΔV.

Worked solution
  1. W = 2.0 × 10−6 × 50.
  2. W = 1.0 × 10−4 J.
Q3Warm-up
How are equipotential surfaces related to electric field lines?

Hint

No work is done moving a charge along an equipotential.

Worked solution
  1. Moving along an equipotential, ΔV = 0, so the field does no work.
  2. So the field must be at right angles to the equipotential: B.

Exam-style questions

Questions at the level of the AP Physics C: E&M exam. The ones marked Challenge combine more than one idea.

Q4Exam-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.
Q5Exam-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.
Q6Exam-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.
Q7Exam-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).
Q8Exam-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.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Mathematical routines style · 6 points · about 20 minutes

A thin ring of radius R carries total charge +Q spread uniformly. Point P is on the ring's axis, a distance z from its centre.

  1. (a) Derive an expression for the electric potential at P.
  2. (b) Use your answer to find the electric field at P.
  3. (c) Explain why the field at the centre of the ring is zero even though the potential is not.

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)

Every bit of charge dq is the same distance √(R2 + z2) from P, so V = kQ/√(R2 + z2).

Part (b)

Ez = −dV/dz = kQz/(R2 + z2)3/2.

Part (c)

At z = 0 the contributions to E from opposite sides of the ring cancel as vectors, but potential is a scalar, so the contributions add to kQ/R.

Your mark: 0 of 6

Common mistakes in electric potential

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics C: E&M past papers, with official scoring guidelines. · All AP Physics C: E&M units