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Conductors and capacitors practice problems AP Physics C: E&M, Unit 10

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

Conductors in electrostatic equilibrium, capacitance, energy stored, dielectrics, and capacitors in series and parallel. 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 conductors and capacitors

C = Q/Vdefinition of capacitance
C = κε0A/dparallel-plate capacitor
U = ½CV2energy stored
Cp = C1 + C2in parallel
1/Cs = 1/C1 + 1/C2in series

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
A 10 μF capacitor is connected to 12 V. What charge does it store, in microcoulombs?
μC

Hint

Q = CV.

Worked solution
  1. Q = 10 μF × 12 V.
  2. Q = 120 μC.
Q2Warm-up
How much energy does that 10 μF capacitor store at 12 V?
J

Hint

U = ½CV2.

Worked solution
  1. U = ½ × 10 × 10−6 × 122.
  2. U = 7.2 × 10−4 J.
Q3Warm-up
Inside a conductor in electrostatic equilibrium, the electric field is:

Hint

In equilibrium, the free charges inside are not moving.

Worked solution
  1. If there were a field inside, free charges would move.
  2. In equilibrium they do not, so the field inside is zero: A.

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
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.
Q5Exam-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.
Q6Exam-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.
Q7Exam-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.)
Q8Challenge
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.

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.

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

Common mistakes in conductors and capacitors

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