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AP Physics C: Electricity and Magnetism Practice Quiz: Conservation of Electric Energy

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

A positive charge is released from rest at a location with high electric potential and is free to move to a location with lower electric potential. According to the principle of conservation of energy, what happens to the charge?

All Questions (9)

A positive charge is released from rest at a location with high electric potential and is free to move to a location with lower electric potential. According to the principle of conservation of energy, what happens to the charge?

A) It gains kinetic energy as its potential energy decreases.

B) It loses kinetic energy as its potential energy increases.

C) It gains both kinetic energy and potential energy.

D) Its kinetic and potential energy remain constant.

Correct Answer: A

A positive charge naturally moves from a region of high potential to a region of low potential. This movement results in a decrease in the system's electric potential energy (ΔU_E < 0). By the conservation of energy, this loss in potential energy is converted into a gain in kinetic energy (ΔK > 0), causing the charge to speed up.

An object with a charge of +2 C is moved through a potential difference of +5 V (from a lower potential to a higher potential). What is the change in the electric potential energy of the object-field system?

A) +10 J

B) -10 J

C) +2.5 J

D) -2.5 J

Correct Answer: A

The change in electric potential energy is given by the equation ΔU_E = qΔV. In this case, the charge q = +2 C and the potential difference ΔV = +5 V. Therefore, ΔU_E = (+2 C)(+5 V) = +10 J.

An electron is moved from a location with an electric potential of 100 V to a location with an electric potential of 20 V. Which of the following correctly describes the changes in energy, assuming only the electric force does work?

A) The system's potential energy increases, and the electron's kinetic energy decreases.

B) The system's potential energy decreases, and the electron's kinetic energy increases.

C) Both the potential energy and kinetic energy increase.

D) Both the potential energy and kinetic energy decrease.

Correct Answer: A

The electron has a negative charge (q < 0). The potential difference is ΔV = V_final - V_initial = 20 V - 100 V = -80 V. The change in potential energy is ΔU_E = qΔV. Since both q and ΔV are negative, their product, ΔU_E, is positive. According to the conservation of energy (ΔK + ΔU_E = 0), if potential energy increases, kinetic energy must decrease by the same amount.

A proton (charge +e) is accelerated from rest by moving through a potential difference of -500 V (i.e., it moves from a higher potential to a lower potential). What is the resulting change in the proton's kinetic energy?

A) A decrease of 500 eV

B) An increase of 500 eV

C) A decrease of 500 J

D) An increase of 500 J

Correct Answer: B

According to the conservation of energy, the change in kinetic energy is the negative of the change in potential energy: ΔK = -ΔU_E. The change in potential energy is ΔU_E = qΔV = (+e)(-500 V) = -500 eV. Therefore, the change in kinetic energy is ΔK = -(-500 eV) = +500 eV, which represents an increase.

A system consists of a fixed arrangement of source charges and a positive test charge. If the positive test charge is moved from a location of low electric potential to a location of high electric potential, how does the electric potential energy of the system change?

A) It increases.

B) It decreases.

C) It remains the same.

D) It becomes zero.

Correct Answer: A

The change in electric potential energy is given by ΔU_E = qΔV. For a positive test charge, q is positive. Moving from a low potential to a high potential means the potential difference, ΔV = V_final - V_initial, is also positive. The product of two positive numbers is positive, so ΔU_E is positive, meaning the potential energy of the system increases.

A particle with charge q = -3 C moves from point A, where the potential is V_A = +10 V, to point B, where the potential is V_B = +20 V. Which statement correctly describes the energy transformation, assuming the particle's speed changes?

A) The system's potential energy increases by 30 J, and the particle's kinetic energy decreases by 30 J.

B) The system's potential energy decreases by 30 J, and the particle's kinetic energy increases by 30 J.

C) The system's potential energy increases by 60 J, and the particle's kinetic energy decreases by 60 J.

D) The system's potential energy decreases by 60 J, and the particle's kinetic energy increases by 60 J.

Correct Answer: B

First, calculate the potential difference: ΔV = V_B - V_A = 20 V - 10 V = +10 V. Next, calculate the change in potential energy: ΔU_E = qΔV = (-3 C)(+10 V) = -30 J. A negative change means the potential energy decreased by 30 J. By the conservation of energy, a decrease in potential energy corresponds to an equal increase in kinetic energy. Therefore, the kinetic energy increases by 30 J.

The existence of a difference in electric potential between two points in space implies that...

A) the electric field must be zero everywhere between the two points.

B) a charged object moving between these points will experience a change in its electric potential energy.

C) any charged object placed between these points will remain stationary.

D) the distance between the two points must be zero.

Correct Answer: B

The provided content directly states that when a charged object moves between two locations with different electric potentials, the electric potential energy of the object-field system changes. This change is a fundamental consequence of the potential difference.

A particle with charge q is moved between two points, resulting in a change in the system's electric potential energy, ΔU_E. If a particle with the same charge q were moved between two new points where the magnitude of the potential difference was doubled, what would be the new change in electric potential energy?

A) 1/2 ΔU_E

B) ΔU_E

C) 2ΔU_E

D) 4ΔU_E

Correct Answer: C

The relationship is ΔU_E = qΔV. Since the change in potential energy, ΔU_E, is directly proportional to the potential difference, ΔV, doubling the potential difference while keeping the charge q constant will result in a doubling of the change in electric potential energy.

An object with a positive charge, +q, moves from point X to point Y. During this movement, the kinetic energy of the object decreases. Which of the following statements about the electric potentials at points X and Y must be true?

A) V_Y > V_X

B) V_Y < V_X

C) V_Y = V_X

D) V_X = 0

Correct Answer: A

According to the conservation of energy, if the kinetic energy decreases (ΔK < 0), the electric potential energy of the system must increase (ΔU_E > 0). The change in potential energy is given by ΔU_E = qΔV. Since the charge q is positive and we know ΔU_E must be positive, the potential difference ΔV = V_Y - V_X must also be positive. A positive potential difference means that the final potential is greater than the initial potential, so V_Y > V_X.