AP Physics C Electricity and Magnetism Quiz: Capacitors

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CapacitorsQuestion 1 of 18

An air-filled parallel-plate capacitor has a capacitance of $C_0$ . If the area of the plates is tripled and the distance between the plates is halved, what is the new capacitance?

\frac{3}{2} C_0

3 C_0

6 C_0

\frac{2}{3} C_0

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Capacitors

Practice Capacitors in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Capacitors, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

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Question 1

An air-filled parallel-plate capacitor has a capacitance of $C_0$ . If the area of the plates is tripled and the distance between the plates is halved, what is the new capacitance?

$\frac{3}{2} C_0$
$3 C_0$
$6 C_0$ (correct answer)
$\frac{2}{3} C_0$

Explanation: The capacitance of a parallel-plate capacitor is given by the formula

C = \frac{\epsilon_0 A}{d}

, where

A

is the area of the plates and

d

is the separation distance. The new area is

A' = 3A

and the new separation is

d' = d/2

. The new capacitance

C'

C' = \frac{\epsilon_0 A'}{d'} = \frac{\epsilon_0 (3A)}{d/2} = 6 \frac{\epsilon_0 A}{d} = 6C_0

Question 2

An electron with initial velocity $v_0$ enters a region of uniform electric field $E$ directed perpendicular to its velocity. The region is between the plates of a large parallel-plate capacitor. Neglecting any edge effects and gravitational forces, which of the following best describes the path of the electron while it is in the electric field?

A straight line in the direction of $v_0$ .
A circular arc.
A parabolic arc. (correct answer)
A straight line in the direction of $E$ .

Explanation: The electron experiences a constant force

F = -eE

in the direction opposite to the electric field. This force is perpendicular to its initial velocity. According to Newton's second law, this results in a constant acceleration

a = F/m_e

in that direction. The situation is analogous to projectile motion under gravity, where there is constant velocity in one direction and constant acceleration in the perpendicular direction. The resulting path is a parabola.

Question 3

A parallel-plate capacitor with charge $Q_0$ is fully discharged by connecting its terminals with a wire of resistance $R$ . The total energy $U_0$ initially stored in the capacitor is dissipated as heat in the wire. If the same capacitor were instead charged to $3Q_0$ , what would be the total energy dissipated in the wire during its discharge?

$U_0/3$
$3U_0$
$9U_0$ (correct answer)
$U_0$

Explanation: The energy stored in a capacitor is given by

U = \frac{Q^2}{2C}

. This is the energy that will be dissipated during discharge. The initial energy with charge

Q_0

U_0 = \frac{Q_0^2}{2C}

. If the capacitor is charged to

3Q_0

, the new stored energy is

U_{new} = \frac{(3Q_0)^2}{2C} = \frac{9Q_0^2}{2C} = 9U_0

. By conservation of energy, this is the total energy that will be dissipated.

Question 4

A parallel-plate capacitor with capacitance $C$ is connected to a battery with a constant potential difference $V$ . While the capacitor is connected to the battery, a dielectric slab with dielectric constant $\kappa$ is inserted between the plates. What is the magnitude of the work done by the battery during this process?

$\frac{1}{2} (\kappa - 1) C V^2$
$(\kappa - 1) C V^2$ (correct answer)
$\frac{1}{2} (1 - 1/\kappa) C V^2$
$(1 - 1/\kappa) C V^2$

Explanation: The work done by the battery is

W_{batt} = (\Delta Q)V

, where

\Delta Q

is the additional charge that flows from the battery. Initially, the charge is

Q_i = CV

. After inserting the dielectric, the capacitance becomes

C_f = \kappa C

, and the final charge is

Q_f = C_f V = \kappa C V

. The additional charge that flows is

\Delta Q = Q_f - Q_i = (\kappa - 1)CV

. Therefore, the work done by the battery is

W_{batt} = (\kappa - 1)CV \cdot V = (\kappa - 1)CV^2

Question 5

A long cylindrical capacitor has an inner conductor of radius $a$ and an outer conductor of radius $b$ . The inner conductor holds a charge per unit length of $+\lambda$ . The electric field in the region $a < r < b$ is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$ . What is the magnitude of the potential difference between the conductors?

$\frac{\lambda}{2\pi\epsilon_0} \ln(b/a)$ (correct answer)
$\frac{\lambda}{2\pi\epsilon_0} (b-a)$
$\frac{\lambda}{2\pi\epsilon_0} (1/a - 1/b)$
$\frac{\lambda b}{2\pi\epsilon_0 a}$

Explanation: The potential difference

V_{ab}

is found by integrating the electric field from the inner to the outer conductor:

V_{ab} = -\int_a^b \vec{E} \cdot d\vec{l}

. Since

E

is directed radially outward, we integrate along a radial path.

V_{ab} = V_a - V_b = \int_a^b E dr = \int_a^b \frac{\lambda}{2\pi\epsilon_0 r} dr = \frac{\lambda}{2\pi\epsilon_0} [\ln(r)]_a^b = \frac{\lambda}{2\pi\epsilon_0} (\ln(b) - \ln(a)) = \frac{\lambda}{2\pi\epsilon_0} \ln(b/a)

Question 6

A parallel-plate capacitor creates a uniform electric field of magnitude $E$ in the volume between its plates. The permittivity of free space is $\epsilon_0$ . What is the energy density, or energy stored per unit volume, in the electric field?

$\frac{1}{2}\epsilon_0 E^2$ (correct answer)
$\epsilon_0 E^2$
$\frac{E^2}{2\epsilon_0}$
$\frac{\epsilon_0 E}{2}$

Explanation: The energy stored in a capacitor is

U = \frac{1}{2}CV^2

. For a parallel-plate capacitor,

C = \epsilon_0 A/d

and

V = Ed

. Substituting these gives

U = \frac{1}{2}(\frac{\epsilon_0 A}{d})(Ed)^2 = \frac{1}{2}\epsilon_0 A d E^2

. The volume between the plates is

Vol = Ad

. The energy density

u_E

U/Vol = \frac{\frac{1}{2}\epsilon_0 A d E^2}{Ad} = \frac{1}{2}\epsilon_0 E^2

Question 7

A parallel-plate capacitor is connected to a battery with a constant potential difference $V_0$ . While connected, the area of the plates that directly overlaps is decreased by a factor of two. What is the ratio of the final stored energy to the initial stored energy?

$1/4$
$1/2$ (correct answer)
$2$
$4$

Explanation: While connected to the battery, the potential difference

V_0

remains constant. The initial energy is

U_i = \frac{1}{2}C_i V_0^2

. The capacitance

C

is proportional to the plate area

A

. Halving the area halves the capacitance, so

C_f = C_i/2

. The final energy is

U_f = \frac{1}{2}C_f V_0^2 = \frac{1}{2}(\frac{C_i}{2})V_0^2 = \frac{1}{2} U_i

. The ratio

U_f/U_i

1/2

Question 8

The energy stored in the electric field of a charged system can be calculated by integrating the energy density $u_E = \frac{1}{2} \epsilon_0 E^2$ over all space. For an isolated conducting sphere of radius $R$ and charge $Q$ , the electric field for $r > R$ is $E = \frac{Q}{4\pi\epsilon_0 r^2}$ . Which integral correctly calculates the energy stored in the field outside the sphere?

$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$ (correct answer)
$\int_0^R \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$
$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right) (4\pi r^2) dr$
$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 dr$

Explanation: To find the total energy, we must integrate the energy density

u_E

over the volume of space where the field exists. The volume element

dV

for a spherical shell of radius

r

and thickness

dr

dV = 4\pi r^2 dr

. The integration must be performed over the region outside the sphere, so the limits are from

r=R

r=\infty

. The integrand is

u_E dV

. Therefore, the correct integral is

U = \int_{V} u_E dV = \int_R^{\infty} \frac{1}{2} \epsilon_0 E^2 (4\pi r^2) dr = \int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr

Question 9

A capacitor with capacitance $C$ is charged to a potential difference $V$ , storing an amount of energy $U$ . If the potential difference is increased to $3V$ , what is the new amount of energy stored in the capacitor?

$U/3$
$3U$
$6U$
$9U$ (correct answer)

Explanation: The energy

U

stored in a capacitor is given by

U = \frac{1}{2}C V^2

. Since the energy is proportional to the square of the potential difference, tripling the voltage from

V

3V

will increase the stored energy by a factor of

3^2 = 9

. The new energy will be

9U

Question 10

An isolated parallel-plate capacitor has plates of area $A$ separated by a distance $d$ . It holds a net charge of magnitude $Q$ on each plate. What is the magnitude of the electrostatic force exerted by one plate on the other?

$\frac{Q^2}{\epsilon_0 A}$
$\frac{Q^2}{2\epsilon_0 A}$ (correct answer)
$\frac{Q^2 d}{\epsilon_0 A}$
$\frac{Q^2}{2\epsilon_0 A d}$

Explanation: The electric field produced by one plate alone is

E_1 = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2\epsilon_0 A}

. The force exerted by this field on the other plate, which has charge

Q

, is

F = Q E_1 = Q \left(\frac{Q}{2\epsilon_0 A}\right) = \frac{Q^2}{2\epsilon_0 A}

. Using the total field between the plates (

E = \sigma/\epsilon_0

) is incorrect because a plate does not exert a force on itself.

Question 11

A coaxial cylindrical capacitor has a length $L$ , an inner conductor of radius $a$ , and an outer conductor of radius $b$ . Its capacitance is given by $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$ . If the inner radius $a$ is halved and the outer radius $b$ is doubled, how does the new capacitance $C'$ compare to the original capacitance $C$ ?

$C' = \frac{\ln(b/a)}{\ln(4b/a)} C$ (correct answer)
$C' = \frac{\ln(b/a)}{2\ln(b/a) + \ln(4)} C$
$C' = \frac{1}{4} C$
$C' = 4 C$

Explanation: The original capacitance is

C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}

. The new radii are

a' = a/2

and

b' = 2b

. The new capacitance is

C' = \frac{2\pi\epsilon_0 L}{\ln(b'/a')} = \frac{2\pi\epsilon_0 L}{\ln((2b)/(a/2))} = \frac{2\pi\epsilon_0 L}{\ln(4b/a)}

. The ratio

C'/C = \frac{\ln(b/a)}{\ln(4b/a)}

. Therefore,

C' = \frac{\ln(b/a)}{\ln(4b/a)} C

Question 12

A spherical capacitor consists of two concentric conducting spheres of radii $a$ and $b$ , with $b>a$ . Its capacitance is given by $C = 4\pi\epsilon_0 \frac{ab}{b-a}$ . What expression is obtained for the capacitance in the limit that the outer sphere's radius $b$ approaches infinity, representing an isolated sphere of radius $a$ ?

$4\pi\epsilon_0 a$ (correct answer)
$4\pi\epsilon_0 b$
$4\pi\epsilon_0 (b-a)$
Zero

Explanation: To find the limit as

b \to \infty

, we can rewrite the expression as

C = 4\pi\epsilon_0 \frac{a}{1 - a/b}

. As

b \to \infty

, the term

a/b

approaches 0. Therefore, the expression for the capacitance approaches

C = 4\pi\epsilon_0 \frac{a}{1 - 0} = 4\pi\epsilon_0 a

. This is the capacitance of an isolated conducting sphere of radius

a

Question 13

An uncharged capacitor is connected to a battery. The process of charging involves moving a total charge $Q_f$ from one plate to the other, resulting in a final potential difference $V_f$ . The total work required to charge the capacitor is $W$ . What is the average potential difference through which each infinitesimal amount of charge is moved during the charging process?

$V_f$
$V_f/2$ (correct answer)
$2V_f$
$W/V_f$

Explanation: The work done to charge a capacitor is equal to the energy stored,

W = U = \frac{1}{2}Q_f V_f

. The work can also be thought of as the total charge moved multiplied by the average potential difference,

W = Q_f V_{avg}

. Equating the two expressions for work gives

Q_f V_{avg} = \frac{1}{2}Q_f V_f

. Solving for the average potential difference yields

V_{avg} = V_f/2

. This reflects that the potential difference increases linearly from 0 to

V_f

as the capacitor is charged.

Question 14

A parallel-plate capacitor with capacitance $C_0$ is charged to potential difference $V_0$ and then disconnected from the battery. A dielectric slab with dielectric constant $\kappa > 1$ is then inserted, completely filling the space between the plates. How does the final stored energy $U_f$ compare to the initial stored energy $U_i$ ?

$U_f = \kappa U_i$
$U_f = U_i / \kappa$ (correct answer)
$U_f = \kappa^2 U_i$
$U_f = U_i / \kappa^2$

Explanation: Since the capacitor is disconnected from the battery, the charge

Q_0

on its plates remains constant. The initial energy is

U_i = Q_0^2 / (2C_0)

. When the dielectric is inserted, the capacitance becomes

C_f = \kappa C_0

. The final energy is

U_f = Q_0^2 / (2C_f) = Q_0^2 / (2\kappa C_0) = (1/\kappa) (Q_0^2 / (2C_0)) = U_i / \kappa

. Since

\kappa > 1

, the stored energy decreases.

Question 15

A parallel-plate capacitor is charged by a battery to store energy $U_0$ . The capacitor is then disconnected from the battery. The distance between the plates is then doubled. What is the new energy $U_f$ stored in the capacitor?

$U_0/4$
$U_0/2$
$2U_0$ (correct answer)
$4U_0$

Explanation: When the capacitor is disconnected, the charge

Q

on the plates is conserved. The capacitance

C

is inversely proportional to the plate separation

d

. Doubling

d

halves the capacitance, so

C_f = C_0/2

. The energy stored in a capacitor can be expressed as

U = Q^2/(2C)

. Since

Q

is constant and

C

is halved, the new energy

U_f = Q^2/(2(C_0/2)) = Q^2/C_0 = 2(Q^2/(2C_0)) = 2U_0

Question 16

The electric potential energy $U$ of a uniformly charged conducting sphere of radius $R$ and total charge $Q$ is $U = \frac{Q^2}{8\pi\epsilon_0 R}$ . This can be interpreted as the energy stored in a capacitor. What is the effective self-capacitance of the sphere?

$8\pi\epsilon_0 R$
$4\pi\epsilon_0 R$ (correct answer)
$4\pi\epsilon_0 R^2$
$2\pi\epsilon_0 R$

Explanation: The energy stored in a capacitor is given by the formula

U = \frac{Q^2}{2C}

. We are given the energy of the charged sphere as

U = \frac{Q^2}{8\pi\epsilon_0 R}

. By comparing these two expressions, we can find the capacitance

C

. Setting them equal:

\frac{Q^2}{2C} = \frac{Q^2}{8\pi\epsilon_0 R}

. Solving for

C

gives

2C = 8\pi\epsilon_0 R

, so

C = 4\pi\epsilon_0 R

. This corresponds to the capacitance of an isolated sphere.

Question 17

A parallel-plate capacitor is held with its plates horizontal. It is charged and then disconnected from the battery. A slab of dielectric material is placed on the bottom plate and released. The slab is pulled into the region between the plates. Which of the following is a correct explanation for this phenomenon?

The dielectric is pulled in because the capacitor does work on it, which increases the capacitor's stored electrical energy.
The dielectric is pulled in to a position of lower potential energy, which occurs because inserting it increases the capacitance. (correct answer)
The magnetic field created by the charge on the plates interacts with the dielectric, causing an attractive force.
The dielectric material becomes polarized, with a net charge that is attracted to the nearest plate and repelled by the farther plate, resulting in a net repulsive force.

Explanation: The system moves to a state of lower potential energy. For an isolated capacitor, the energy is

U = Q^2/(2C)

. When the dielectric is inserted, the capacitance

C

increases. Since

Q

is constant, the stored energy

U

decreases. The decrease in potential energy is converted into the kinetic energy of the slab and the work done by the electric field, indicating an attractive force that pulls the slab in.

Question 18

The electric field between the plates of a charged parallel-plate capacitor is nearly uniform, but there are non-uniform 'fringing fields' near the edges. The formula $C = \epsilon_0 A/d$ is derived assuming a uniform field. How does the actual capacitance of a real capacitor compare to the value predicted by this formula?

The actual capacitance is greater, because the fringing fields store additional energy. (correct answer)
The actual capacitance is smaller, because the fringing fields weaken the field in the central region.
The actual capacitance is the same, because the effects of the fringing fields cancel out over the entire volume.
The actual capacitance is smaller, because the fringing fields cause charge to leak from the edges.

Explanation: Capacitance is defined as

C=Q/V

. For a given charge

Q

, the fringing fields mean that the electric field lines occupy a larger effective volume, extending beyond the edges of the plates. This effectively stores more energy for the same amount of charge compared to the ideal case. A greater storage capacity for a given charge implies a greater capacitance. Equivalently, for a given voltage V, more charge can be stored due to the fringing fields, thus increasing capacitance.