AP Physics 2: Algebra-Based Practice Quiz: Electric Potential
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 15 questions to check your progress.
Question 1 of 15
All Questions (15)
A) The force experienced by a charge at a point in space.
B) The total kinetic energy of a charged particle.
C) The electric potential energy per unit charge.
D) The work required to create a configuration of charges.
Correct Answer: C
The provided content explicitly states: 'Electric potential describes the electric potential energy per unit charge at a point in space.'
A) By calculating the vector sum of the individual potentials from each charge.
B) By calculating the scalar sum of the individual potentials from each charge.
C) By considering only the potential from the charge with the greatest magnitude.
D) By finding the average of the maximum and minimum potentials.
Correct Answer: B
The content specifies that 'The electric potential due to multiple point charges can be determined by the principle of scalar superposition of the electric potential due to each of the point charges.' This means the potentials, which are scalar quantities, are added together algebraically.
A) 100 V/m
B) 50 V/m
C) 25 V/m
D) 0.04 V/m
Correct Answer: C
Using the relationship provided, |E| = |ΔV/Δr|. Plugging in the values, |E| = |50 V / 2.0 m| = 25 V/m.
A) The electric field does positive work on the charge.
B) The electric field does negative work on the charge.
C) The electric field does zero work on the charge.
D) The charge accelerates in the direction of the electric field.
Correct Answer: C
Equipotential lines represent paths of constant potential. The work done by the electric field is related to the change in potential energy, W = -qΔV. Since the potential does not change (ΔV = 0) when moving along an equipotential line, the work done by the field is zero. This is an application of the concepts in points 3 and 6.
A) V/3
B) V
C) 3V
D) 9V
Correct Answer: C
According to the equation for the potential of a point charge, V is directly proportional to the charge q ($V=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r}$). If the charge is tripled, the potential will also be tripled. This is a direct application of the superposition principle.
A) The field in X is stronger than in Y.
B) The field in Y is stronger than in X.
C) The fields in X and Y are equal in strength.
D) The field is zero in both regions.
Correct Answer: A
The magnitude of the average electric field is given by |E| = |ΔV/Δr|. A smaller distance Δr for the same potential difference ΔV results in a larger electric field magnitude. Since the lines are closer together in Region X, the electric field is stronger there.
A) +20 V
B) +10 V
C) 0 V
D) -20 V
Correct Answer: C
The principle of scalar superposition states that the total potential is the algebraic sum of the individual potentials. Therefore, V_total = (+10 V) + (-10 V) = 0 V.
A) 1000 V
B) 40 V
C) 10 V
D) 4 V
Correct Answer: C
Using the relationship |ΔV| = |E| * |Δr|. First, convert the distance to meters: 5 cm = 0.05 m. Then, calculate the potential difference: |ΔV| = (200 N/C) * (0.05 m) = 10 V.
A) Joule per meter (J/m)
B) Volt per meter (V/m)
C) Coulomb per second (C/s)
D) Watt per Coulomb (W/C)
Correct Answer: B
The equation directly relates the electric field (E) to the potential difference (ΔV), measured in Volts, and the distance (Δr), measured in meters. Therefore, the unit for the electric field can be expressed as Volts per meter (V/m).
A) higher electric potential.
B) lower electric potential.
C) zero electric field.
D) the same electric potential.
Correct Answer: B
The electric field vector points in the direction of the steepest decrease in electric potential. A positive charge experiences a force in the direction of the electric field. Therefore, a positive charge will move from a region of higher potential to a region of lower potential.
A) 20 J
B) 40 J
C) 80 J
D) 0.05 J
Correct Answer: C
Electric potential (V) is defined as electric potential energy (U_E) per unit charge (q), so V = U_E / q. Rearranging for potential energy gives U_E = qV. Therefore, U_E = (2 C) * (40 V) = 80 J.
A) Magnetic field maps and gravitational potential lines
B) Electric current loops and resistance diagrams
C) Electric field vector maps and equipotential lines
D) Force diagrams and kinetic energy graphs
Correct Answer: C
The provided content states: 'Electric field vector maps and equipotential lines are tools to describe the field produced by a charge or configuration of charges and can be used to predict the motion of charged objects in the field.'
A) (a, 0)
B) (-2a, 0)
C) (4a, 0)
D) (-a, 0)
Correct Answer: B
The total potential is V = kQ/r1 + k(-2Q)/r2. We need V=0, so kQ/r1 = 2kQ/r2, which simplifies to r2 = 2r1. Let the point be (x,0). r1 = |x| and r2 = |x-2a|. If x = -2a, then r1 = |-2a| = 2a and r2 = |-2a - 2a| = |-4a| = 4a. Here, r2 = 2r1, so the potential is zero. For the other options: A) r1=a, r2=a -> V≠0. C) r1=4a, r2=2a -> V≠0. D) r1=a, r2=3a -> V≠0.
A) only the direction of the charges matters.
B) the direction from the charge to the point of interest must be considered using vectors.
C) the potentials add algebraically without concern for direction in space.
D) only positive potentials contribute to the total.
Correct Answer: C
As a scalar quantity, electric potential has magnitude but no direction. The principle of scalar superposition means that the individual potentials are simply added together, taking their positive or negative signs into account, to find the total potential.
A) Zero
B) k * q / r
C) 4 * k * q / r
D) 4 * k * q / s
Correct Answer: C
Using the principle of scalar superposition, the total potential is the sum of the potentials from each of the four charges. Since each charge +q is the same distance r from the center, the total potential is V_total = V1 + V2 + V3 + V4 = (k*q/r) + (k*q/r) + (k*q/r) + (k*q/r) = 4 * k * q / r. (Here k = 1/(4πε₀)).