AP Physics C Electricity and Magnetism Quiz: Electric Potential Energy

20 questions · exam conditions

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Electric Potential EnergyQuestion 1 of 20

A charge of $+q$ and a charge of $-2q$ are located on the x-axis at $x=0$ and $x=d$ , respectively. How much work must an external agent do to bring a third charge of $+3q$ from infinity to the point $x=2d$ on the x-axis?

-k \frac{3q^2}{d}

-k \frac{9q^2}{2d}

+k \frac{3q^2}{2d}

+k \frac{q^2}{d}

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Potential Energy

Practice Electric Potential Energy 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 Electric Potential Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$-k \frac{3q^2}{d}$
$-k \frac{9q^2}{2d}$ (correct answer)
$+k \frac{3q^2}{2d}$
$+k \frac{q^2}{d}$

Explanation: The work done by an external agent is equal to the change in potential energy, which is

W_{ext} = q_{new}V_{final}

, where

V_{final}

is the potential at the destination point due to the existing charges. The potential at

x=2d

due to the charges at

x=0

and

x=d

V_{final} = k \frac{+q}{2d} + k \frac{-2q}{d} = k\left(\frac{q}{2d} - \frac{4q}{2d}\right) = -k\frac{3q}{2d}

. The work done is

W_{ext} = (+3q) V_{final} = (+3q) \left(-k\frac{3q}{2d}\right) = -k\frac{9q^2}{2d}

Question 2

System A consists of two protons separated by a distance $d$ . System B consists of a proton and an electron separated by a distance $d$ . System C consists of two electrons separated by a distance $d$ . Let $U_A$ , $U_B$ , and $U_C$ be the electric potential energies. Which option correctly ranks the potential energies?

$U_A > U_C > U_B$
$U_B > U_A = U_C$
$U_A = U_C > U_B$ (correct answer)
$U_A > U_B > U_C$

Explanation: Let the elementary charge be

e

. The potential energies are:

U_A = k \frac{(+e)(+e)}{d} = +k \frac{e^2}{d}

U_B = k \frac{(+e)(-e)}{d} = -k \frac{e^2}{d}

U_C = k \frac{(-e)(-e)}{d} = +k \frac{e^2}{d}

. Since

k, e, d

are positive,

U_A

and

U_C

are equal and positive, while

U_B

is negative. Therefore, the correct ranking from most positive to most negative is

U_A = U_C > U_B

Question 3

The electric potential energy of a system consisting of a point charge $q$ and a fixed point charge $Q$ is $U(r) = kQq/r$ . Which of the following describes the work done by the electric field as the charge $q$ is moved from an initial distance $R$ to a final distance $3R$ ?

$-k \frac{2Qq}{3R}$
$+k \frac{2Qq}{3R}$ (correct answer)
$-k \frac{Qq}{3R}$
$+k \frac{Qq}{3R}$

Explanation: The work done by the conservative electric field is the negative of the change in potential energy:

W_E = -\Delta U = -(U_f - U_i)

. The initial potential energy is

U_i = kQq/R

. The final potential energy is

U_f = kQq/(3R)

. The change in potential energy is

\Delta U = U_f - U_i = \frac{kQq}{3R} - \frac{kQq}{R} = \frac{kQq - 3kQq}{3R} = -\frac{2kQq}{3R}

. The work done by the field is

W_E = -(-\frac{2kQq}{3R}) = +\frac{2kQq}{3R}

Question 4

A proton is released from rest in a region of space with a non-uniform electric field. As the proton moves from point A to point B, its kinetic energy increases by $5.0 \times 10^{-16} \text{ J}$ . What is the change in the electric potential energy of the proton-field system?

$+5.0 \times 10^{-16} \text{ J}$
$-5.0 \times 10^{-16} \text{ J}$ (correct answer)
Zero, because energy is conserved.
The change cannot be determined without knowing the path taken.

Explanation: According to the principle of conservation of energy for the proton-field system, the total energy remains constant. Since only the conservative electric force does work, the sum of the changes in kinetic energy (

\Delta K

) and electric potential energy (

\Delta U_E

) must be zero.

\Delta K + \Delta U_E = 0

. Given that the kinetic energy increases by

5.0 \times 10^{-16} \text{ J}

, we have

\Delta K = +5.0 \times 10^{-16} \text{ J}

. Therefore,

\Delta U_E = -\Delta K = -5.0 \times 10^{-16} \text{ J}

Question 5

A particle of mass $m$ and charge $-q$ is projected with an initial speed $v_0$ directly away from a fixed stationary particle of charge $+Q$ . The initial separation is $r_0$ . What is the minimum initial speed $v_0$ required for the particle to escape to an infinite distance from the fixed particle?

$\sqrt{\frac{kQq}{mr_0}}$
$\sqrt{\frac{2kQq}{mr_0}}$ (correct answer)
$\sqrt{\frac{kQq}{2mr_0}}$
$\frac{kQq}{mr_0}$

Explanation: For the particle to just escape, its total mechanical energy must be zero. The total energy at infinity (

r=\infty

) is zero (

K_f = 0, U_f = 0

). By conservation of energy, the initial total energy must also be zero.

E_i = K_i + U_i = 0

. The initial kinetic energy is

K_i = \frac{1}{2}mv_0^2

. The initial potential energy is

U_i = k \frac{(+Q)(-q)}{r_0} = -\frac{kQq}{r_0}

. Setting the sum to zero:

\frac{1}{2}mv_0^2 - \frac{kQq}{r_0} = 0

. Solving for

v_0

gives

v_0 = \sqrt{\frac{2kQq}{mr_0}}

Question 6

A system consists of four charges fixed at the corners of a square of side $s$ . The charges are $+q$ , $-q$ , $+q$ , and $-q$ , arranged such that adjacent charges have opposite signs. What is the total electric potential energy of this configuration?

$k \frac{q^2}{s}(4 - \sqrt{2})$
$k \frac{q^2}{s}(\sqrt{2} - 4)$ (correct answer)
$k \frac{q^2}{s}(2 - 2\sqrt{2})$
Zero

Explanation: The total potential energy is the sum of the potential energies of all six unique pairs. There are four adjacent pairs (distance

s

) and two diagonal pairs (distance

s\sqrt{2}

). Each adjacent pair consists of opposite charges (

+q, -q

), so there are four such terms:

4 \times k\frac{(+q)(-q)}{s} = -4k\frac{q^2}{s}

. The two diagonal pairs consist of like charges (

+q, +q

and

-q, -q

k\frac{(+q)(+q)}{s\sqrt{2}} + k\frac{(-q)(-q)}{s\sqrt{2}} = \frac{2kq^2}{s\sqrt{2}} = \frac{\sqrt{2}kq^2}{s}

. The total energy is

U_{total} = -4k\frac{q^2}{s} + \sqrt{2}k\frac{q^2}{s} = k \frac{q^2}{s}(\sqrt{2} - 4)

Question 7

An electron is accelerated from rest through a potential difference of $V$ . It then enters a region of uniform electric field where it is brought to rest. The change in the electron's electric potential energy during the initial acceleration phase is $\Delta U_1$ , and during the deceleration phase is $\Delta U_2$ . Which of the following is correct?

$\Delta U_1 > 0$ , $\Delta U_2 < 0$ , and $|\Delta U_1| = |\Delta U_2|$
$\Delta U_1 < 0$ , $\Delta U_2 > 0$ , and $|\Delta U_1| = |\Delta U_2|$ (correct answer)
$\Delta U_1 < 0$ , $\Delta U_2 < 0$ , and $|\Delta U_1| = |\Delta U_2|$
$\Delta U_1 > 0$ , $\Delta U_2 > 0$ , and $|\Delta U_1| > |\Delta U_2|$

Explanation: During acceleration from rest, the electron's kinetic energy increases. By conservation of energy, its potential energy must decrease. So,

\Delta U_1 < 0

. The change is

\Delta U_1 = (-e)V

. During deceleration to rest, its kinetic energy decreases back to zero. Thus, its potential energy must increase, so

\Delta U_2 > 0

. By the work-energy theorem, the total change in kinetic energy is zero, so the net work done by the electric field is zero. This means the total change in potential energy is zero:

\Delta U_1 + \Delta U_2 = 0

, which implies

\Delta U_2 = -\Delta U_1

. Therefore, their magnitudes are equal.

Question 8

In the context of the electric field, calculate the change in electric potential energy for a $-2.0\,\mu\text{C}$ charge moving from 5 V to 25 V.

$+4.0\times10^{-5}\,\text{J}$
$-4.0\times10^{-5}\,\text{J}$ (correct answer)
$+20\,\text{V}$
$-40\,\text{J}$

Explanation: This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the negative charge moving from 5 V to 25 V experiences a potential difference of ΔV = 25 - 5 = 20 V. The correct answer, B, is derived by applying ΔU = qΔV = (-2.0×10⁻⁶ C)(20 V) = -4.0×10⁻⁵ J. A common distractor, A, arises from forgetting that negative charges lose potential energy when moving to higher potentials. To assist students, emphasize that the sign of ΔU depends on both the charge sign and the direction of potential change, and practice interpreting physical meaning of energy changes.

Question 9

Based on the scenario, calculate the change in electric potential energy for a $+6.0\,\mu\text{C}$ charge moved through a $-8\,\text{V}$ potential difference.

$-4.8\times10^{-5}\,\text{J}$ (correct answer)
$+4.8\times10^{-5}\,\text{J}$
$-8\,\text{J}$
$-4.8\times10^{-6}\,\text{J}$

Explanation: This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge experiences a potential difference of ΔV = -8 V (given directly). The correct answer, A, is derived by applying ΔU = qΔV = (6.0×10⁻⁶ C)(-8 V) = -4.8×10⁻⁵ J. A common distractor, B, arises from misunderstanding that a negative potential difference means the charge moves to a lower potential. To assist students, emphasize that potential difference signs indicate direction of change, and positive charges lose energy when experiencing negative potential differences.

Question 10

Three identical point charges, each with charge $+q$ , are fixed at the vertices of an equilateral triangle with side length $s$ . What is the total electric potential energy of this system of charges?

$k \frac{q^2}{s}$
$2k \frac{q^2}{s}$
$3k \frac{q^2}{s}$ (correct answer)
$\sqrt{3} k \frac{q^2}{s}$

Explanation: The total electric potential energy of a system of charges is the scalar sum of the potential energies of all unique pairs of charges. In this system, there are three unique pairs of charges. Since all charges are

+q

and all separation distances are

s

, the potential energy of each pair is

U_{pair} = k \frac{q \cdot q}{s} = k \frac{q^2}{s}

. The total potential energy is the sum for the three pairs:

U_{total} = 3 \times U_{pair} = 3k \frac{q^2}{s}

Question 11

An external agent moves a $-3.0 \text{ C}$ point charge at constant velocity from a point with an electric potential of $+20 \text{ V}$ to a point with an electric potential of $-10 \text{ V}$ . What is the work done on the charge by the external agent?

$-90 \text{ J}$
$-30 \text{ J}$
$+30 \text{ J}$
$+90 \text{ J}$ (correct answer)

Explanation: The work done by an external agent to move a charge at constant velocity is equal to the change in the electric potential energy of the charge:

W_{ext} = \Delta U_E = q \Delta V = q(V_{final} - V_{initial})

. Here,

q = -3.0 \text{ C}

V_{initial} = +20 \text{ V}

, and

V_{final} = -10 \text{ V}

. So,

W_{ext} = (-3.0 \text{ C})(-10 \text{ V} - 20 \text{ V}) = (-3.0 \text{ C})(-30 \text{ V}) = +90 \text{ J}

Question 12

A point charge $+q$ is fixed at the origin. An external force moves a second point charge $+Q$ from position A at ( $r_0, 0$ ) to position B at ( $0, r_0$ ). What is the work done by the electric field of the fixed charge during this displacement?

$k \frac{qQ}{r_0}$
$-k \frac{qQ}{r_0}$
Zero (correct answer)
$\sqrt{2} k \frac{qQ}{r_0}$

Explanation: The work done by the conservative electric field is equal to the negative of the change in electric potential energy:

W_E = -\Delta U_E = -(U_B - U_A)

. The electric potential energy of the system when the charge

+Q

is at a distance

r

from

+q

U = k \frac{qQ}{r}

. Both position A and position B are at the same distance,

r_0

, from the origin. Therefore,

U_A = U_B = k \frac{qQ}{r_0}

. The change in potential energy

\Delta U_E

is zero, and thus the work done by the electric field is also zero.

Question 13

The electric potential in a region of space is described by the function $V(x, y) = 4xy^2$ , where $V$ is in volts and $x$ and $y$ are in meters. An external agent moves a proton (charge $e = 1.6 \times 10^{-19} \text{ C}$ ) from the origin ( $0,0$ ) to the point ( $2 \text{ m}, 1 \text{ m}$ ) at a constant velocity. What is the change in the electric potential energy of the proton?

$1.28 \times 10^{-18} \text{ J}$ (correct answer)
$2.56 \times 10^{-18} \text{ J}$
$5.12 \times 10^{-18} \text{ J}$
$1.02 \times 10^{-17} \text{ J}$

Explanation: The change in electric potential energy is given by

\Delta U_E = q \Delta V = q(V_f - V_i)

. The initial potential at the origin (

0,0

) is

V_i = 4(0)(0)^2 = 0 \text{ V}

. The final potential at (

2 \text{ m}, 1 \text{ m}

) is

V_f = 4(2)(1)^2 = 8 \text{ V}

. The change in potential is

\Delta V = 8 \text{ V} - 0 \text{ V} = 8 \text{ V}

. The change in potential energy is

\Delta U_E = (1.6 \times 10^{-19} \text{ C})(8 \text{ V}) = 12.8 \times 10^{-19} \text{ J} = 1.28 \times 10^{-18} \text{ J}

Question 14

A system in a stable equilibrium configuration is disturbed by a small displacement. Which statement must be true about the system's electric potential energy and the work done by the conservative electric forces within the system?

The potential energy increases, and the electric forces do negative work. (correct answer)
The potential energy decreases, and the electric forces do positive work.
The potential energy remains constant, and the electric forces do zero work.
The potential energy increases, and the electric forces do positive work.

Explanation: A stable equilibrium corresponds to a local minimum in the potential energy function. When the system is slightly displaced from this minimum, its potential energy must increase. The work done by a conservative force, such as the electric force, is equal to the negative of the change in potential energy:

W_E = -\Delta U_E

. Since the potential energy increases (

\Delta U_E > 0

), the work done by the electric forces must be negative. This negative work corresponds to a restoring force that tends to bring the system back to equilibrium.

Question 15

Three point charges $+Q$ , $-Q$ , and $+2Q$ are placed at the vertices of an equilateral triangle of side length $s$ . An external agent disassembles the system, moving each charge to an infinite distance from the others. What is the total work done by the external agent?

$k \frac{Q^2}{s}$ (correct answer)
$k \frac{2Q^2}{s}$
$-k \frac{Q^2}{s}$
$k \frac{3Q^2}{s}$

Explanation: The work done by an external agent to disassemble the system is

W_{ext} = U_{final} - U_{initial}

. The final potential energy, with all charges at infinite separation, is

U_{final} = 0

. The initial potential energy is the sum of the energies of the three pairs:

U_{initial} = k\frac{(+Q)(-Q)}{s} + k\frac{(+Q)(+2Q)}{s} + k\frac{(-Q)(+2Q)}{s} = k\frac{-Q^2+2Q^2-2Q^2}{s} = -k\frac{Q^2}{s}

. The work done is

W_{ext} = 0 - \left(-k\frac{Q^2}{s}\right) = +k\frac{Q^2}{s}

Question 16

The work required to assemble a system of four identical positive charges $+q$ at the corners of a square of side $s$ is $W_A$ . The work required to assemble a system of four identical positive charges $+q$ at the corners of a square of side $2s$ is $W_B$ . Which statement correctly compares the work done?

$W_A = 4 W_B$
$W_A = 2 W_B$ (correct answer)
$W_A = W_B$
$W_A = W_B / 2$

Explanation: The work to assemble a configuration is equal to its total electric potential energy. The potential energy is the sum of the potential energies of all pairs. For a square of side length

L

, there are 4 pairs at distance

L

and 2 pairs at distance

L\sqrt{2}

. The total energy is

U = 4k\frac{q^2}{L} + 2k\frac{q^2}{L\sqrt{2}} = \frac{kq^2}{L}(4 + \sqrt{2})

. The work is proportional to

1/L

. Thus,

W_A \propto 1/s

and

W_B \propto 1/(2s)

. This means

W_B = W_A / 2

, or

W_A = 2W_B

Question 17

An isolated system consists of two particles with charges $+q$ and $+4q$ and masses $m$ and $4m$ , respectively. They are released from rest when separated by a distance $d$ . What is the ratio of the kinetic energy of charge $+q$ to the kinetic energy of charge $+4q$ long after their release?

1:16
1:4
1:1
4:1 (correct answer)

Explanation: The system is isolated, so momentum is conserved. Since they start from rest, the total momentum is always zero. Therefore, their momenta must be equal in magnitude and opposite in direction:

|p_1| = |p_2|

, so

mv_1 = (4m)v_2

, which gives

v_1 = 4v_2

. The kinetic energy is

K = \frac{1}{2}mv^2

. The ratio of their kinetic energies is

\frac{K_1}{K_2} = \frac{\frac{1}{2}m v_1^2}{\frac{1}{2}(4m)v_2^2} = \frac{1}{4} \left(\frac{v_1}{v_2}\right)^2

. Substituting

v_1 = 4v_2

, we get

\frac{K_1}{K_2} = \frac{1}{4} (4)^2 = \frac{16}{4} = 4

. The ratio is 4:1.

Question 18

In the context of the electric field, calculate the change in electric potential energy for a $+1.0\,\mu\text{C}$ charge moved 0.50 m across a uniform $200\,\text{V/m}$ field.

$+1.0\times10^{-4}\,\text{J}$
$-1.0\times10^{-4}\,\text{J}$ (correct answer)
$+100\,\text{V}$
$+2.0\times10^{-4}\,\text{J}$

Explanation: This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the charge moved 0.50 m across a 200 V/m field, experiencing a potential difference of ΔV = -Ed = -200 × 0.50 = -100 V (negative because moving along field direction). The correct answer, B, is derived by applying ΔU = qΔV = (1.0×10⁻⁶ C)(-100 V) = -1.0×10⁻⁴ J. A common distractor, A, arises from sign confusion about field direction. To assist students, emphasize that "across" typically means along the field direction (high to low potential), and practice visualizing field lines and equipotential surfaces.

Question 19

A system consists of a proton and an electron separated by a finite distance. Which of the following best describes the electric potential energy of this system?

Positive, because the charges are elementary and fundamental.
Negative, because work must be done on the system to separate the particles to an infinite distance. (correct answer)
Zero, because the net charge of the system is zero, making it electrically neutral overall.
Negative, because the system will spontaneously release energy as the particles are separated to an infinite distance.

Explanation: The electric potential energy of two point charges is

U_E = k \frac{q_1 q_2}{r}

. For a proton (

q_1 = +e

) and an electron (

q_2 = -e

), the product

q_1 q_2

is negative, so the potential energy is negative. A negative potential energy for a bound system implies that energy must be added (positive work must be done by an external force) to overcome the attractive force and separate the components to a state of zero potential energy (infinite separation).

Question 20

Based on the scenario, calculate the change in electric potential energy for a $+1.5\,\mu\text{C}$ charge moved through a battery from 0 V to 12 V.

$+1.8\times10^{-5}\,\text{J}$ (correct answer)
$+18\,\text{J}$
$+12\,\text{J}$
$-1.8\times10^{-5}\,\text{J}$

Explanation: This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge moved through a battery from 0 V to 12 V experiences a potential difference of ΔV = 12 - 0 = 12 V. The correct answer, A, is derived by applying ΔU = qΔV = (1.5×10⁻⁶ C)(12 V) = 1.8×10⁻⁵ J. A common distractor, B, arises from incorrect unit conversion, treating microCoulombs as Coulombs. To assist students, emphasize careful unit conversion (1 μC = 10⁻⁶ C) and practice dimensional analysis to ensure energy units (Joules) result from charge (Coulombs) times potential (Volts).