AP Physics 2 Quiz: Electric Potential Energy

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 2.

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.

Question 1

1. increases because the charge moves to a higher electric potential.
2. decreases because the charge moves to a higher electric potential. (correct answer)
3. remains the same because electric potential is independent of charge.
4. decreases because a negative charge always loses potential energy when it moves.

Explanation: Electric potential energy. The electric potential energy relationship U = qV shows that for a negative charge (-3.0 μC), the potential energy is negative when at positive potential. Moving from V_C = +5 V to V_D = +25 V, the change in potential energy is ΔU = q(V_f - V_i) = (-3.0 μC)(+25 V - 5 V) = -60 μJ, indicating a decrease. A negative charge at higher positive potential has more negative (lower) potential energy, similar to how a negatively charged particle is repelled from positive regions. Choice D incorrectly assumes negative charges always lose energy when moving, without considering the potential difference. Track both charge sign and potential when reasoning about electric potential energy.

Question 2

1. It increases because the charge moves to a lower electric potential.
2. It decreases because the charge moves to a lower electric potential. (correct answer)
3. It remains the same because only the electric field strength matters.
4. It increases because a positive charge always gains potential energy when it moves.

Explanation: Electric potential energy. The electric potential energy of a charge is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC), moving from higher potential (+30 V) to lower potential (+10 V) means ΔU = q(V_f - V_i) = (+2.0 μC)(+10 V - 30 V) = -40 μJ, indicating a decrease in potential energy. When a positive charge moves to lower potential, it moves in the direction the electric field would naturally push it, like a ball rolling downhill, thus losing potential energy. Choice D incorrectly assumes positive charges always gain energy when moving, ignoring that the direction relative to potential matters. Track both charge sign and potential when reasoning about electric potential energy.

Question 3

1. increases because a negative charge gains energy when it moves.
2. decreases because the charge is negative.
3. remains the same because the electric potential is the same at both points. (correct answer)
4. increases because electric potential energy equals electric potential.

Explanation: Electric potential energy. When a charge moves between two points at the same potential (V_L = V_M = +12 V), the change in potential energy is ΔU = q(V_f - V_i) = (-1.0 μC)(+12 V - 12 V) = 0 μJ. The potential energy remains constant because U = qV depends only on the potential at a location, not on the path taken or the sign of the charge when potentials are equal. This is true regardless of whether the charge is positive or negative. Choice D incorrectly equates electric potential energy with electric potential, failing to recognize that U = qV includes the charge as a factor. Track both charge sign and potential when reasoning about electric potential energy.

Question 4

1. It decreases because the charge is negative and the potential is higher.
2. It increases because the charge is negative and the potential increases.
3. It stays the same because only the electric field strength matters.
4. It decreases because the charge is negative and the potential increases. (correct answer)

Explanation: This problem tests understanding of electric potential energy. Electric potential energy is calculated as U = qV, where q is the charge and V is the electric potential. For a negative charge (-3.0 μC) moving from V = +5 V to V = +25 V, the change in potential energy is ΔU = q(Vf - Vi) = (-3.0 μC)(+25 V - 5 V) = (-3.0 μC)(+20 V) = -60 μJ. Since ΔU is negative, the potential energy decreases. Choice B incorrectly claims energy increases, likely confusing the fact that negative charges naturally move toward higher potential with the energy change. Remember: track both charge sign and potential when reasoning about electric potential energy.

Question 5

1. decreases because the charge moves to a higher electric potential (correct answer)
2. increases because the charge moves to a higher electric potential
3. increases because the charge moves to a lower electric potential
4. remains the same because both potentials are negative

Explanation: This question tests understanding of electric potential energy. Electric potential energy is U = qV, where q is the charge and V is the potential at that point. For a negative charge (-0.80 μC) moving from V_E = -50 V to V_F = -20 V, we calculate ΔU = q(V_F - V_E) = (-0.80 μC)(-20 V - (-50 V)) = (-0.80 μC)(+30 V) = -24 μJ. Since ΔU is negative, the potential energy decreases as the negative charge moves to higher (less negative) potential. Choice C incorrectly assumes that both potentials being negative means no change, ignoring the actual calculation. Track both charge sign and potential when reasoning about electric potential energy: ΔU = qΔV.

Question 6

1. increase because the charge is positive and the potential increases
2. remain the same because only the electric field strength matters
3. decrease because the charge is positive and the potential decreases (correct answer)
4. increase because the charge is positive and the potential decreases

Explanation: This problem tests understanding of electric potential energy. Electric potential energy is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC), the initial energy is U_A = (+2.0 μC)(+80 V) = +160 μJ, and the final energy is U_B = (+2.0 μC)(+20 V) = +40 μJ. Since U_B < U_A, the potential energy decreases by 120 μJ. Choice A incorrectly assumes that positive charges gain energy when moving to lower potential, confusing this with the behavior of negative charges. When solving electric potential energy problems, remember that U = qV directly gives the energy, and positive charges naturally move toward lower potential, losing potential energy in the process.

Question 7

1. decreases because the charge moves to a higher electric potential
2. decreases because the charge moves to a lower electric potential (correct answer)
3. increases because the charge moves to a lower electric potential
4. remains the same because the charge is released from rest

Explanation: This question tests understanding of electric potential energy. Electric potential energy is given by U = qV, where q is the charge and V is the potential. For a positive charge (+4.0 μC) moving from V_M = +60 V to V_N = +20 V, we calculate ΔU = q(V_N - V_M) = (+4.0 μC)(+20 V - 60 V) = (+4.0 μC)(-40 V) = -160 μJ. Since ΔU is negative, the potential energy decreases as the positive charge moves to lower potential. Choice C incorrectly assumes that being released from rest affects the potential energy change, when initial motion state doesn't change the energy calculation. Track potential energy changes using ΔU = qΔV, remembering positive charges lose potential energy moving to lower potentials.

Question 8

1. decreases because the charge moves to a higher electric potential
2. increases because the charge moves to a lower electric potential
3. remains the same because only the electric field strength matters
4. decreases because the charge moves to a lower electric potential (correct answer)

Explanation: This question tests understanding of electric potential energy. Electric potential energy is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC) moving from V_A = +30 V to V_B = -10 V, we calculate the change: ΔU = q(V_B - V_A) = (+2.0 μC)(-10 V - 30 V) = (+2.0 μC)(-40 V) = -80 μJ. Since ΔU is negative, the potential energy decreases. Choice C incorrectly assumes that only electric field strength matters, ignoring that potential energy depends on both charge and potential difference. When solving potential energy problems, always calculate ΔU = qΔV and check the sign to determine if energy increases or decreases.

Question 9

1. decreases because the electric field strength must be smaller at location 2
2. remains constant because potential is a property of space, not energy
3. increases because the charge moves to a higher electric potential (correct answer)
4. decreases because the charge moves to a higher electric potential

Explanation: This question tests understanding of electric potential energy. The electric potential energy of a charge is U = qV, where q is the charge and V is the potential at that location. For a positive charge (+1.5 μC) moving from V_1 = -20 V to V_2 = +10 V, we find ΔU = q($V_2$ - $V_1$) = (+1.5 μC)(+10 V - (-20 V)) = (+1.5 μC)(+30 V) = +45 μJ. Since ΔU is positive, the potential energy increases. Choice D incorrectly relates potential energy change to electric field strength, when potential energy depends only on charge and potential difference. To solve potential energy problems, use ΔU = qΔV and remember that positive charges gain energy moving to higher potentials.

Question 10

1. remains the same because the potential at A is zero
2. decreases because the charge moves to a higher electric potential (correct answer)
3. increases because the charge moves to a higher electric potential
4. decreases because the charge moves to a lower electric potential

Explanation: This question tests understanding of electric potential energy. Electric potential energy equals U = qV, where q is the charge and V is the electric potential. For a negative charge (-1.0 μC) moving from V_A = 0 V to V_B = +40 V, the change is ΔU = q(V_B - V_A) = (-1.0 μC)(+40 V - 0 V) = (-1.0 μC)(+40 V) = -40 μJ. Since ΔU is negative, the potential energy decreases as the negative charge moves to higher potential. Choice C incorrectly assumes zero potential means no energy change, failing to recognize that potential differences drive energy changes. Always calculate ΔU = qΔV to determine energy changes, regardless of whether one potential is zero.

Question 11

1. increases because $\Delta U=q\Delta V>0$ (correct answer)
2. decreases because the potential becomes more negative
3. decreases because $\Delta U=\Delta V/q$
4. remains the same because the charge is negative

Explanation: Electric potential energy. The electric potential energy is U = qV, and when a charge moves, the energy change is ΔU = qΔV. For this negative charge (-3.0 μC) moving from 0 V to -50 V, the potential change is ΔV = -50 V - 0 V = -50 V. With both q and ΔV negative, we get ΔU = (-3.0 μC)(-50 V) = +150 μJ, which is positive, indicating an increase in potential energy. Choice A incorrectly assumes that moving to more negative potential always decreases energy, failing to account for the charge sign - negative charges gain energy when moving to lower (more negative) potentials. Track both charge sign and potential when reasoning about electric potential energy.

Question 12

1. It increases because the final potential has a larger magnitude.
2. It stays the same because potential energy depends only on charge sign.
3. It increases because the charge is positive and the potential becomes negative.
4. It decreases because the charge is positive and the potential decreases. (correct answer)

Explanation: This problem tests understanding of electric potential energy. Electric potential energy is U = qV, where q is the charge and V is the electric potential. For a positive charge (+9.0 nC) moving from VX = +3 V to VY = -9 V, the change in potential energy is ΔU = q(Vf - Vi) = (+9.0 nC)(-9 V - 3 V) = (+9.0 nC)(-12 V) = -108 nJ. Since ΔU is negative, the potential energy decreases. Choice C incorrectly focuses on the magnitude of the final potential being larger, ignoring that -9 V is actually lower than +3 V. Remember: always consider the algebraic sign of potentials when calculating energy changes.

Question 13

1. It increases because the charge is positive and the potential decreases.
2. It stays the same because the field is external and unchanged.
3. It increases because electric potential is lower at the final point.
4. It decreases because the charge is positive and the potential decreases. (correct answer)

Explanation: This problem tests understanding of electric potential energy. Electric potential energy is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC) moving from V = +30 V to V = +10 V, the change in potential energy is ΔU = q(Vf - Vi) = (+2.0 μC)(+10 V - 30 V) = (+2.0 μC)(-20 V) = -40 μJ. Since ΔU is negative, the potential energy decreases. Choice C incorrectly suggests energy increases just because the final potential is lower, ignoring that a positive charge naturally moves toward lower potential and loses potential energy. Remember: for positive charges, moving to lower potential means decreasing potential energy.

Question 14

1. increase because the charge is positive
2. change because the path length affects potential energy
3. decrease because the charge is positive
4. remain the same because the initial and final potentials are equal (correct answer)

Explanation: This question tests understanding of electric potential energy as a state function. Since electric potential energy depends only on position (through U = qV), not on the path taken, and both initial and final potentials are 120 V, we have U_E = (+3.0 μC)(120 V) = +360 μJ and U_F = (+3.0 μC)(120 V) = +360 μJ. Therefore, ΔU = 0 and the energy remains the same. Choice D incorrectly suggests that path length matters for potential energy, confusing this with work done against friction or other non-conservative forces. Remember that electric potential energy is path-independent; only the initial and final positions (potentials) determine the energy change.

Question 15

1. decreases because the electric field does work along the path
2. increases because the charge is positive
3. remains the same because the electric field strength must be zero
4. remains the same because the electric potential is the same at R and S (correct answer)

Explanation: This question tests understanding of electric potential energy. Electric potential energy is U = qV, and for a charge moving between two points, the change is ΔU = q(V_final - V_initial). Here, a positive charge (+0.50 μC) moves from V_R = -12 V to V_S = -12 V, so ΔU = (+0.50 μC)(-12 V - (-12 V)) = (+0.50 μC)(0 V) = 0 μJ. Since the potentials are equal, there is no change in potential energy. Choice B incorrectly assumes the electric field does work along any path, but in electrostatics, work depends only on endpoints, not path. Remember that electric potential energy change depends only on the potential difference between endpoints: ΔU = qΔV.

Question 16

1. decreases because electric potential increased, independent of charge sign
2. decreases because $\Delta U=q\Delta V<0$ (correct answer)
3. increases because the charge moves to higher potential
4. remains the same because the charge magnitude is constant

Explanation: Electric potential energy. The potential energy of a charge in an electric field is U = qV, and the change in potential energy when moving between points is ΔU = qΔV. For this negative charge (-2.0 μC) moving from V_P = 10 V to V_Q = 70 V, the change in potential is ΔV = 70 V - 10 V = +60 V. Therefore, ΔU = (-2.0 μC)(+60 V) = -120 μJ, which is negative, indicating a decrease in potential energy. Choice A incorrectly assumes that moving to higher potential always increases energy, failing to account for the negative charge which experiences the opposite effect. Track both charge sign and potential when reasoning about electric potential energy.

Question 17

1. increase because the potential increases
2. decrease because the charge is negative and the potential increases (correct answer)
3. increase because the charge is negative and the potential increases
4. remain the same because $0\,\text{V}$ means zero energy

Explanation: This problem examines electric potential energy when a negative charge moves to zero potential. Using U = qV, the initial energy is U_G = (-2.5 μC)(-100 V) = +250 μJ, and the final energy is U_H = (-2.5 μC)(0 V) = 0 μJ. Since the energy changes from +250 μJ to 0 μJ, the potential energy decreases by 250 μJ. Choice D incorrectly assumes that V = 0 means U = 0 for all charges, forgetting that U = qV depends on both charge and potential. When dealing with zero potential, remember it's just a reference point; the actual energy U = qV can be positive, negative, or zero depending on the charge.

Question 18

1. remains the same because $\Delta V\neq 0$ only changes potential
2. increases because $\Delta U=q\Delta V>0$ (correct answer)
3. decreases because the charge moved to lower potential
4. increases because negative charges gain energy when potential decreases

Explanation: Electric potential energy. The electric potential energy of a charge-field system is U = qV, and changes according to ΔU = qΔV when the charge moves. For this negative charge (-1.0 nC) moving from 200 V to 50 V, the potential change is ΔV = 50 V - 200 V = -150 V. With a negative charge and negative ΔV, we calculate ΔU = (-1.0 nC)(-150 V) = +150 nJ, which is positive, indicating an increase in energy. Choice D contains the correct physical insight but uses incorrect terminology - negative charges do gain energy when moving to lower potential, but this is because ΔU = qΔV becomes positive when both q and ΔV are negative. Track both charge sign and potential when reasoning about electric potential energy.

Question 19

1. decreases because only the distance moved determines $\Delta U$
2. increases because $\Delta U=q\Delta V>0$ (correct answer)
3. remains the same because $V_M$ is zero
4. decreases because the charge is positive

Explanation: Electric potential energy. The potential energy of a charge in an electric field is U = qV, where q is the charge and V is the potential. When a charge moves between points, the energy change is ΔU = qΔV = q(V_N - V_M). For this positive charge (+6.0 nC) moving from 0 V to 120 V, we have ΔV = 120 V - 0 V = +120 V, so ΔU = (+6.0 nC)(+120 V) = +720 nJ, which is positive. Choice C incorrectly assumes that because the initial potential is zero, the energy remains unchanged, missing that what matters is the change in potential, not whether one value is zero. Track both charge sign and potential when reasoning about electric potential energy.

Question 20

1. increases because the charge moves to a lower electric potential. (correct answer)
2. decreases because the charge moves to a lower electric potential.
3. remains the same because both potentials are negative.
4. decreases because electric potential equals electric potential energy.

Explanation: Electric potential energy. A negative charge (-0.80 μC) moving from V_U = -3 V to V_V = -18 V sees the potential become more negative (decrease). The change in potential energy is ΔU = q(V_f - V_i) = (-0.80 μC)(-18 V - (-3 V)) = (-0.80 μC)(-15 V) = +12 μJ. The positive change indicates an increase in potential energy because negative charges have higher potential energy at lower (more negative) potentials. Choice C incorrectly assumes that both potentials being negative means no change, ignoring the actual difference in values. Track both charge sign and potential when reasoning about electric potential energy.