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AP Physics 2: Algebra-Based 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 charged particle moves from a location with one electric potential to a different location with another electric potential. According to the principles of energy conservation, which of the following quantities for the particle-field system must change?

All Questions (9)

A charged particle moves from a location with one electric potential to a different location with another electric potential. According to the principles of energy conservation, which of the following quantities for the particle-field system must change?

A) The mass of the particle

B) The sign of the particle's charge

C) The electric potential energy

D) The total energy of the system

Correct Answer: C

The provided content states that a difference in electric potential between two locations leads to changes in energy in the system. Specifically, it defines the change in electric potential energy as ΔUE = qΔV. Since there is a difference in potential (ΔV ≠ 0), there must be a change in electric potential energy.

A particle with charge +q moves through an electric potential difference ΔV, and the electric potential energy of the system changes by an amount ΔUE. If a second particle with charge +2q moves through the same potential difference ΔV, what is the change in electric potential energy of the new system?

A) 1/2 ΔUE

B) ΔUE

C) 2 ΔUE

D) 4 ΔUE

Correct Answer: C

The change in electric potential energy is given by the equation ΔUE = qΔV. The change in potential energy is directly proportional to the charge q. If the charge is doubled from q to 2q while the potential difference ΔV remains the same, the new change in potential energy will be (2q)ΔV, which is equal to 2(qΔV) or 2ΔUE.

A charged object moves between two points in an electric field. If the electric potential energy of the object-field system decreases during this movement, what is the corresponding change in the object's kinetic energy, assuming no other forces act on the object?

A) The kinetic energy increases.

B) The kinetic energy decreases.

C) The kinetic energy remains constant.

D) The kinetic energy becomes zero.

Correct Answer: A

This question is based on the conservation of energy. The total energy of the system (kinetic + potential) is conserved. If the electric potential energy (UE) decreases, the kinetic energy (K) must increase to keep the total energy constant. This is stated in the content: 'The movement...results in a change in kinetic energy...consistent with the conservation of energy.'

A positive charge is released from rest and moves from a region of high electric potential to a region of low electric potential. Which of the following correctly describes the changes in energy?

A) Electric potential energy increases, and kinetic energy decreases.

B) Electric potential energy decreases, and kinetic energy increases.

C) Both electric potential energy and kinetic energy increase.

D) Both electric potential energy and kinetic energy decrease.

Correct Answer: B

For a positive charge (q > 0), moving from a high potential to a low potential means the potential difference ΔV is negative. Using ΔUE = qΔV, a positive q multiplied by a negative ΔV results in a negative ΔUE, meaning electric potential energy decreases. By the conservation of energy, a decrease in potential energy corresponds to an increase in kinetic energy.

A negative charge moves from a region of high electric potential to a region of low electric potential. Which of the following correctly describes the changes in energy?

A) Electric potential energy increases, and kinetic energy decreases.

B) Electric potential energy decreases, and kinetic energy increases.

C) Both electric potential energy and kinetic energy increase.

D) Both electric potential energy and kinetic energy decrease.

Correct Answer: A

For a negative charge (q < 0), moving from a high potential to a low potential means the potential difference ΔV is negative. Using ΔUE = qΔV, a negative q multiplied by a negative ΔV results in a positive ΔUE, meaning electric potential energy increases. By the conservation of energy, an increase in potential energy corresponds to a decrease in kinetic energy.

A particle with charge q is accelerated from rest through a potential difference ΔV, gaining a final kinetic energy K. What would be the final kinetic energy of a particle with charge q but twice the mass, if it were also accelerated from rest through the same potential difference ΔV?

A) 1/2 K

B) K

C) 2 K

D) 4 K

Correct Answer: B

According to the principle of conservation of energy, the kinetic energy gained (ΔK) is equal to the magnitude of the electric potential energy lost (-ΔUE). The change in potential energy is given by ΔUE = qΔV. Therefore, the kinetic energy gained is K = -qΔV. This relationship depends only on the charge (q) and the potential difference (ΔV), not on the mass of the particle. Since both q and ΔV are the same for the second particle, its final kinetic energy will also be K.

The change in the electric potential energy of an object-field system as the object moves between two locations is determined by which two properties?

A) The object's mass and the distance traveled.

B) The object's charge and the potential difference between the locations.

C) The object's initial velocity and its mass.

D) The potential difference between the locations and the time taken to move.

Correct Answer: B

The provided content explicitly gives the equation for the change in electric potential energy as ΔUE = qΔV. This equation shows that the change in energy depends directly on the object's charge (q) and the electric potential difference (ΔV) between the two locations.

A charged particle moves between two points in space, and its kinetic energy is observed to decrease. What can be definitively concluded about the particle's charge (q) and the electric potential difference it moved through (ΔV)?

A) The charge q must be negative, and the potential difference ΔV must be positive.

B) The charge q and the potential difference ΔV must have the same sign (both positive or both negative).

C) The charge q and the potential difference ΔV must have opposite signs.

D) The potential difference ΔV must be zero.

Correct Answer: B

If the kinetic energy decreases, then by conservation of energy, the electric potential energy must increase. This means ΔUE is a positive value. From the equation ΔUE = qΔV, for the product of two numbers (q and ΔV) to be positive, they must have the same sign. Either both are positive (a positive charge moving to a higher potential) or both are negative (a negative charge moving to a lower potential).

A negatively charged particle is observed to increase in speed as it moves from point A to point B. Which statement correctly describes the electric potentials at points A and B?

A) The potential at A is higher than the potential at B (VA > VB).

B) The potential at B is higher than the potential at A (VB > VA).

C) The potentials at A and B are equal (VA = VB).

D) The potential at A must be zero.

Correct Answer: B

An increase in speed means the kinetic energy increases (ΔK > 0). By conservation of energy, the electric potential energy must decrease (ΔUE < 0). The particle has a negative charge (q < 0). Using the equation ΔUE = qΔV, for ΔUE to be negative when q is negative, the potential difference ΔV = VB - VA must be positive. A positive ΔV means that the final potential (VB) is higher than the initial potential (VA).