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AP Physics C: Electricity and Magnetism Practice Quiz: Capacitors

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

According to the provided description, which of the following best describes the physical properties of a parallel-plate capacitor?

All Questions (10)

According to the provided description, which of the following best describes the physical properties of a parallel-plate capacitor?

A) A single conducting sphere.

B) Two parallel conducting plates separated by a distance.

C) A conducting wire coiled into a solenoid.

D) Two concentric conducting cylinders.

Correct Answer: B

The content explicitly describes a parallel-plate capacitor based on its physical properties, which consist of two parallel conducting plates. The other options describe different electrical components or configurations not mentioned in the text.

The capacitance of a capacitor is a measure of its ability to store charge for a given potential difference. Based on the equation $C=\frac{Q}{\Delta V}$, what does capacitance represent?

A) The total charge a capacitor can hold.

B) The ratio of the charge stored on a plate to the potential difference between the plates.

C) The potential difference created by a certain amount of charge.

D) The energy stored per unit of charge.

Correct Answer: B

The equation $C=\frac{Q}{\Delta V}$ directly defines capacitance (C) as the ratio of the magnitude of the charge (Q) on each plate to the electric potential difference (ΔV) between the plates.

A parallel-plate capacitor holds a charge of magnitude Q when the potential difference between its plates is ΔV. If the charge is increased to 3Q, what is the new potential difference across the plates, assuming the capacitance remains constant?

A) \frac{1}{3}\Delta V

B) \Delta V

C) 3\Delta V

D) 9\Delta V

Correct Answer: C

According to the capacitance equation, $C=\frac{Q}{\Delta V}$, which can be rearranged to $\Delta V = \frac{Q}{C}$. Since the capacitance C is a constant property of the capacitor, the potential difference ΔV is directly proportional to the charge Q. If the charge is tripled (from Q to 3Q), the potential difference must also triple to 3ΔV.

The electric field between the plates of a parallel-plate capacitor is given by $E=\frac{Q}{\epsilon_{0}A}$. If the charge Q on the plates is doubled and the area A of the plates is halved, how does the new electric field E' compare to the original field E?

A) E' = \frac{1}{4}E

B) E' = E

C) E' = 2E

D) E' = 4E

Correct Answer: D

The original electric field is $E=\frac{Q}{\epsilon_{0}A}$. The new charge is Q' = 2Q and the new area is A' = A/2. The new electric field is $E'=\frac{Q'}{\epsilon_{0}A'} = \frac{2Q}{\epsilon_{0}(A/2)} = 4\frac{Q}{\epsilon_{0}A} = 4E$.

The equation for the magnitude of the electric field between two charged parallel plates is given as $E=\frac{Q}{\epsilon_{0}A}$, with the condition that the plate separation is much smaller than the dimensions of the plates. What does this equation imply about the electric field?

A) The electric field is stronger near the positive plate.

B) The electric field is dependent on the material between the plates.

C) The electric field is uniform and does not depend on the distance between the plates.

D) The electric field is inversely proportional to the charge on the plates.

Correct Answer: C

The variables in the equation $E=\frac{Q}{\epsilon_{0}A}$ are the charge (Q), the plate area (A), and the permittivity of free space (ε₀). The distance (separation) between the plates is not a variable in this equation. This implies that for a given charge and plate area, the electric field is uniform (has the same magnitude) at all points between the plates.

A capacitor stores an amount of electric potential energy $U_C$ when a charge of magnitude Q is on its plates, creating a potential difference of ΔV. If the potential difference is kept constant but the charge stored is doubled, what is the new stored energy?

A) \frac{1}{2}U_C

B) U_C

C) 2U_C

D) 4U_C

Correct Answer: C

The electric potential energy stored in a capacitor is given by $U_{C}=\frac{1}{2}Q\Delta V$. If the charge Q is doubled to 2Q while the potential difference ΔV remains constant, the new energy $U'_{C}$ will be $U'_{C}=\frac{1}{2}(2Q)\Delta V = 2(\frac{1}{2}Q\Delta V) = 2U_C$.

A given parallel-plate capacitor stores energy $U_C$ when it holds a charge of magnitude Q. If the charge on the plates is increased to 2Q, what is the new amount of energy stored in the capacitor?

A) \frac{1}{2}U_C

B) U_C

C) 2U_C

D) 4U_C

Correct Answer: D

First, use $C=\frac{Q}{\Delta V}$ to find the effect on potential difference. For a given capacitor, C is constant. So, $\Delta V = \frac{Q}{C}$. If Q is doubled to 2Q, the new potential difference is $\Delta V' = \frac{2Q}{C} = 2\Delta V$. Now, use the energy equation $U_{C}=\frac{1}{2}Q\Delta V$. The new energy is $U'_{C}=\frac{1}{2}Q'\Delta V' = \frac{1}{2}(2Q)(2\Delta V) = 4(\frac{1}{2}Q\Delta V) = 4U_C$.

A capacitor stores an amount of energy $U_C$ when the potential difference across its plates is $\Delta V$. How much charge Q is stored on the plates?

A) Q = \frac{2U_C}{\Delta V}

B) Q = \frac{U_C}{2\Delta V}

C) Q = \sqrt{2U_C C}

D) Q = \frac{1}{2}U_C \Delta V

Correct Answer: A

The question asks to express the charge Q in terms of the stored energy $U_C$ and potential difference $\Delta V$. Starting with the equation for stored energy, $U_{C}=\frac{1}{2}Q\Delta V$. To solve for Q, multiply both sides by 2 and divide by $\Delta V$, which yields $Q = \frac{2U_C}{\Delta V}$.

The magnitude of the electric field between the plates of a certain capacitor is E when it holds a charge Q. If the charge is reduced to Q/3 and the plate area is doubled, what is the magnitude of the new electric field, E'?

A) E' = \frac{1}{6}E

B) E' = \frac{2}{3}E

C) E' = \frac{3}{2}E

D) E' = 6E

Correct Answer: A

The electric field is given by $E=\frac{Q}{\epsilon_{0}A}$. The new charge is Q' = Q/3 and the new area is A' = 2A. Substituting these into the equation gives the new field: $E' = \frac{Q'}{\epsilon_{0}A'} = \frac{Q/3}{\epsilon_{0}(2A)} = \frac{1}{6} \frac{Q}{\epsilon_{0}A} = \frac{1}{6}E$.

A parallel-plate capacitor with constant capacitance C initially holds charge Q. This results in an electric field E and stored potential energy U. The charge on the capacitor is then increased to 3Q. What are the new electric field E' and new stored potential energy U'?

A) E' = 3E and U' = 3U

B) E' = 3E and U' = 9U

C) E' = 9E and U' = 3U

D) E' = 9E and U' = 9U

Correct Answer: B

First, analyze the electric field: $E=\frac{Q}{\epsilon_{0}A}$. Since E is directly proportional to Q, if the charge becomes 3Q, the new electric field will be E' = 3E. Next, analyze the energy. For a fixed capacitor, from $C=\frac{Q}{\Delta V}$, the potential difference $\Delta V$ is proportional to Q. So, if Q is tripled, $\Delta V$ is also tripled ($\\Delta V' = 3\Delta V$). The new stored energy is $U'=\frac{1}{2}Q'\Delta V' = \frac{1}{2}(3Q)(3\Delta V) = 9(\frac{1}{2}Q\Delta V) = 9U$.