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
All Questions (11)
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.
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) 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.
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) 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) +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.
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.
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) 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) 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) 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.