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AP Physics 2: Algebra-Based 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 11 questions to check your progress.

Question 1 of 11

Which of the following best describes the fundamental structure of a parallel-plate capacitor based on the provided text?

All Questions (11)

Which of the following best describes the fundamental structure of a parallel-plate capacitor based on the provided text?

A) A single conducting sphere designed to hold charge.

B) Two separated parallel conducting surfaces.

C) A long solenoid wire wrapped around a core.

D) Two conducting plates that must be in physical contact.

Correct Answer: B

Content point 2 states, 'A parallel-plate capacitor consists of two separated parallel conducting surfaces that can hold equal amounts of charge with opposite signs.' This directly describes the physical structure.

According to the definition provided, capacitance relates which two physical quantities?

A) The area of the plates and the distance between them.

B) The stored potential energy and the charge on a plate.

C) The magnitude of the charge on a plate and the electric potential difference between the plates.

D) The dielectric constant and the permittivity of free space.

Correct Answer: C

Content point 3 states, 'Capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by the separation of those charges,' which is represented by the equation C = Q / ΔV.

A parallel-plate capacitor has a certain capacitance. If the area of its plates is doubled, while all other factors remain constant, how does the capacitance change?

A) It is halved.

B) It remains the same.

C) It is doubled.

D) It is quadrupled.

Correct Answer: C

According to the equation C = κε₀(A/d), capacitance (C) is directly proportional to the area (A) of the plates. Therefore, doubling the area will double the capacitance.

How would the capacitance of a parallel-plate capacitor be affected if the distance between its plates were to be doubled, assuming all other properties are unchanged?

A) The capacitance would be halved.

B) The capacitance would be doubled.

C) The capacitance would be quartered.

D) The capacitance would remain the same.

Correct Answer: A

The equation C = κε₀(A/d) shows that capacitance (C) is inversely proportional to the distance (d) between the plates. If the distance is doubled, the capacitance will be halved.

A capacitor is connected to a battery, resulting in a charge of 12 μC on each plate and a potential difference of 3 V across the plates. What is the capacitance of this capacitor?

A) 36 μF

B) 4 μF

C) 0.25 μF

D) 9 μF

Correct Answer: B

Using the definition of capacitance, C = Q / ΔV. Plugging in the given values: C = 12 μC / 3 V = 4 μF.

A parallel-plate capacitor is designed to hold equal amounts of charge with opposite signs on its two conducting surfaces. If the plate connected to the positive terminal of a battery holds a charge of +Q, what is the charge on the other plate?

A) +Q

B) 0

C) +2Q

D) -Q

Correct Answer: D

Content point 2 explicitly states that a parallel-plate capacitor holds 'equal amounts of charge with opposite signs.' Therefore, if one plate has a charge of +Q, the other must have a charge of -Q.

The electric potential energy stored in a capacitor is given by the equation U_C = (1/2)QΔV. If a capacitor holds a charge of 4 C at a potential difference of 10 V, how much energy is stored?

A) 40 J

B) 20 J

C) 8 J

D) 1.25 J

Correct Answer: B

Using the formula for stored energy, U_C = (1/2)QΔV. Substituting the values: U_C = (1/2)(4 C)(10 V) = 20 J.

The geometry of a parallel-plate capacitor is modified such that the area of its plates is tripled and the distance between the plates is also tripled. What is the effect on its capacitance?

A) The capacitance is multiplied by 9.

B) The capacitance is multiplied by 3.

C) The capacitance remains unchanged.

D) The capacitance is divided by 3.

Correct Answer: C

The original capacitance is C = κε₀(A/d). The new capacitance is C' = κε₀(3A/3d). The factors of 3 in the numerator and denominator cancel out, so C' = κε₀(A/d) = C. The capacitance remains unchanged.

A given parallel-plate capacitor with capacitance C is connected to a variable power source. The voltage is increased, causing the charge stored on the plates to double. What is the new capacitance of the capacitor?

A) C/2

B) C

C) 2C

D) 4C

Correct Answer: B

The capacitance of a capacitor is a physical property determined by its geometry (Area and distance) and the dielectric material between its plates (C = κε₀A/d). It does not depend on the charge stored or the potential difference applied. Therefore, even if the charge doubles, the capacitance C remains constant.

A capacitor stores energy U_C when it has a charge Q and a potential difference ΔV. If the charge is doubled to 2Q and the potential difference is halved to ΔV/2, what is the new stored energy in terms of U_C?

A) U_C / 4

B) U_C / 2

C) U_C

D) 2 * U_C

Correct Answer: C

The original energy is U_C = (1/2)QΔV. The new energy, U'_C, is calculated with the new values: U'_C = (1/2)(2Q)(ΔV/2). The factors of 2 and 1/2 cancel, leaving U'_C = (1/2)QΔV, which is equal to the original energy U_C.

A parallel-plate capacitor is connected to a battery that maintains a constant potential difference ΔV. The capacitor initially stores energy U_C. If the area of the plates is then doubled while it remains connected to the battery, what is the new energy stored in the capacitor?

A) U_C / 2

B) U_C

C) 2 * U_C

D) 4 * U_C

Correct Answer: C

This is a multi-step problem. First, determine the change in capacitance: since C is proportional to A (from C = κε₀A/d), doubling the area doubles the capacitance to 2C. Second, determine the new charge: since the capacitor stays connected to the battery, ΔV is constant. From Q = CΔV, the new charge Q' = (2C)ΔV = 2Q. Finally, calculate the new energy: U'_C = (1/2)Q'ΔV = (1/2)(2Q)ΔV = 2 * [(1/2)QΔV] = 2 * U_C.