---
title: "AP Physics 2 Unit 10 Practice Test"
description: "A timed AP Physics 2 Unit 10 (Electric Force, Field, and Potential) practice test in the AP format, drawn from 1,424 multiple-choice questions and 35 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-physics-2-revised/unit-test/10"
type: "unit-test"
subject: "AP Physics 2"
unit: "Unit 10"
---

# AP Physics 2 Unit 10 Practice Test

## Overview

A timed AP Physics 2 Unit 10 (Electric Force, Field, and Potential) practice test in the AP format, drawn from 1,424 multiple-choice questions and 35 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

## Unit FRQs by Type

- [FRQ 1 – Mathematical Routines](/ap-physics-2-revised/frqs/mathematical-routines?unit=10): 7 questions
- [FRQ 2 – Translation Between Representations](/ap-physics-2-revised/frqs/translation-between-representations?unit=10): 7 questions
- [FRQ 3 – Experimental Design](/ap-physics-2-revised/frqs/experimental-design?unit=10): 9 questions
- [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-2-revised/frqs/qualitative-quantitative-translation?unit=10): 12 questions

## Practice Questions

### Question 1

Two point charges, $$+Q$$ and $$-Q$$, are fixed on the x-axis at $$x = -d$$ and $$x = +d$$, respectively. Which statement correctly describes the electric potential at the origin ($$x = 0$$)?

- (A) The potential is zero because the scalar contributions from the positive and negative charges cancel out
- (B) The potential is zero because the electric field vectors from the two charges point in opposite directions
- (C) The potential is nonzero because the scalar contributions from the two charges add up constructively
- (D) The potential is nonzero because the electric field vectors from the two charges point in the same direction

**Answer: (A)** Applying the principle of superposition requires summing the scalar potentials from each charge ($$V = kQ/d + k(-Q)/d = 0$$). The correct answer recognizes that electric potential is a scalar quantity where equal positive and negative contributions at the same distance sum to zero.

### Question 2

A conducting sphere of radius $$R$$ with an initial charge $$+Q$$ is connected by a thin wire to an uncharged conducting sphere of radius $$3R$$. Once equilibrium is reached, what is the final charge on the larger sphere?

- (A) The final charge is $$3Q/4$$ because charge flows until the electric potentials of both spheres are equal
- (B) The final charge is $$Q/2$$ because charge flows until the charge magnitudes on both spheres are equal
- (C) The final charge is $$Q/4$$ because charge flows until the surface electric fields of both spheres are equal
- (D) The final charge is $$Q/3$$ because charge flows until the electric potential energies of both spheres are equal

**Answer: (A)** Applying the condition for electrostatic equilibrium between conductors requires equating their potentials ($$V = kq/r$$). Since the larger sphere has 3 times the radius, it must hold 3 times the charge ($$q_{large} = 3q_{small}$$) to have the same potential, resulting in $$3/4$$ of the total charge $$Q$$.

### Question 3

Four point charges, each with magnitude $$+Q$$, are fixed at the corners of a square with side length $$d$$. Which expression correctly represents the total electric potential at the center of the square?

- (A) The potential is $$4\sqrt{2}kQ/d$$ because the scalar potentials from all four charges add together
- (B) The potential is $$0$$ because the electric field vectors from all four charges cancel out
- (C) The potential is $$4kQ/d$$ because the distance to the center is treated as $$d$$ for each charge
- (D) The potential is $$2\sqrt{2}kQ/d$$ because only diagonal pairs contribute to the total potential

**Answer: (A)** Calculating the potential requires summing the scalar contribution from each charge ($$V = \sum kq/r$$). The distance from each corner to the center is $$d/\sqrt{2}$$, so the total potential is $$4 \times (kQ / (d/\sqrt{2})) = 4\sqrt{2}kQ/d$$.

### Question 4

A positive point charge $$+Q$$ is fixed at the origin. A student measures the electric potential $$V$$ and electric field magnitude $$E$$ at a distance $$r$$ from the charge. If the student moves to a distance $$2r$$, how do the new values compare?

- (A) The new potential is $$V/2$$ and the new field is $$E/4$$ because $$V \propto 1/r$$ and $$E \propto 1/r^2$$
- (B) The new potential is $$V/4$$ and the new field is $$E/2$$ because $$V \propto 1/r^2$$ and $$E \propto 1/r$$
- (C) The new potential is $$V/2$$ and the new field is $$E/2$$ because both $$V$$ and $$E$$ are proportional to $$1/r$$
- (D) The new potential is $$V/4$$ and the new field is $$E/4$$ because both $$V$$ and $$E$$ are proportional to $$1/r^2$$

**Answer: (A)** Applying the theoretical definitions of potential and field for a point charge reveals their different dependencies on distance. Potential scales inversely with distance ($$1/r$$), while field magnitude scales inversely with the square of distance ($$1/r^2$$).

### Question 5

A point charge $$+Q$$ is placed at $$x=0$$ and a point charge $$-2Q$$ is placed at $$x=D$$. At which position on the x-axis between the charges is the electric potential zero?

- (A) At $$x = D/3$$, because the potential from the positive charge cancels the potential from the negative charge
- (B) At $$x = D/2$$, because the zero potential point is always equidistant from two opposite point charges
- (C) At $$x = 2D/3$$, because the potential is zero closer to the charge with the larger magnitude
- (D) At $$x = D/4$$, because the potential varies with the square of the distance from each charge

**Answer: (A)** Solving for the zero potential point requires setting the sum of potentials to zero: $$kQ/x + k(-2Q)/(D-x) = 0$$. This simplifies to $$1/x = 2/(D-x)$$, which yields $$x = D/3$$, the point where the scalar contributions cancel.

### Question 6

An alpha particle (charge $$+2e$$) moves from a location where the electric potential is $$100 \text{ V}$$ to a location where the potential is $$120 \text{ V}$$ under the influence of only the electric force. Which statement correctly describes the change in the particle's kinetic energy?

- (A) The kinetic energy decreases by $$40 \text{ eV}$$ because the particle moves to a higher potential
- (B) The kinetic energy increases by $$40 \text{ eV}$$ because the electric field does positive work
- (C) The kinetic energy decreases by $$20 \text{ eV}$$ because the potential difference is $$20 \text{ V}$$
- (D) The kinetic energy increases by $$20 \text{ eV}$$ because potential energy is converted to kinetic

**Answer: (A)** Applying conservation of energy, the change in kinetic energy equals the negative change in potential energy ($$\Delta K = -q\Delta V$$). Moving a positive charge to a higher potential increases potential energy ($$+2e \times 20\text{ V} = +40\text{ eV}$$), causing a corresponding decrease in kinetic energy.

### Question 7

A solid conducting sphere of radius $$R$$ carries a net positive charge $$+Q$$ and is in electrostatic equilibrium. Which statement correctly describes the electric potential at a distance $$r = R/2$$ from the center?

- (A) It equals $$kQ/R$$, because the potential is constant throughout a conductor in equilibrium
- (B) It equals $$2kQ/R$$, because the potential increases as the distance to the center decreases
- (C) It equals $$0$$, because the electric field is zero everywhere inside a conductor in equilibrium
- (D) It equals $$kQ/(2R)$$, because the potential decreases as the distance to the center decreases

**Answer: (A)** Applying the properties of conductors requires recognizing that the entire conductor is an equipotential volume. The potential anywhere inside ($$r < R$$) is the same as the potential at the surface ($$kQ/R$$).

### Question 8

Three point charges are arranged at the vertices of an equilateral triangle of side $$s$$. Two vertices have charge $$+q$$ and the third has charge $$-q$$. What is the electric potential at the center of the triangle?

- (A) The potential is $$\sqrt{3}kq/s$$ because the scalar potentials from the three charges sum algebraically
- (B) The potential is $$0$$ because the contributions from the positive and negative charges cancel out
- (C) The potential is $$3\sqrt{3}kq/s$$ because the magnitudes of the potentials from all charges add up
- (D) The potential is $$kq/s$$ because the effective charge at the center is the net charge $$+q$$

**Answer: (A)** Calculating the potential requires summing scalars: $$V = k(q)/r + k(q)/r + k(-q)/r = kq/r$$. The distance $$r$$ from center to vertex is $$s/\sqrt{3}$$, so $$V = kq / (s/\sqrt{3}) = \sqrt{3}kq/s$$.

### Question 9

A battery is connected to a circuit. Which statement best describes the electric potential difference created by the battery in terms of energy and charge?

- (A) It is the work done per unit charge by chemical processes to separate positive and negative charges
- (B) It is the force exerted per unit charge by the chemical processes on the electrons in the terminals
- (C) It is the rate of energy transfer per unit charge as current flows through the internal resistance
- (D) It is the total potential energy stored in the battery divided by the mass of the chemical reactants

**Answer: (A)** Applying the definition of electric potential difference ($$\Delta V = \Delta U_E / q$$) in the context of a battery (LO 10.5.A.3.i) identifies it as the energy provided per unit charge to create charge separation.

### Question 10

Two identical spheres are initially charged: Sphere A has charge $$+4 \mu C$$ and Sphere B has charge $$-2 \mu C$$. They are brought into contact and then separated. Which statement correctly compares their final electric potentials?

- (A) Both spheres have the same positive electric potential because charge redistributes until potentials equalize
- (B) Both spheres have zero electric potential because the positive and negative charges neutralize each other
- (C) Sphere A has a positive potential and Sphere B has a negative potential because they retain their original signs
- (D) Sphere A has a higher potential than Sphere B because it started with a greater magnitude of charge

**Answer: (A)** Applying conservation of charge and the equilibrium condition: total charge is $$+2 \mu C$$, which splits equally ($$+1 \mu C$$ each) between identical spheres. Since both have positive charge and equal radius, they share the same positive potential.

## Review the Unit

- [10.1 Electric Charge and Electric Force](/ap-physics-2-revised/unit-10/1-electric-charge-and-electric-force/study-guide/E6OYkOGeroCXwgw1)
- [10.2 The Process of Charging](/ap-physics-2-revised/unit-10/2-the-process-of-charging/study-guide/fQJqVklZDhbQdyBa)
- [10.3 Electric Fields](/ap-physics-2-revised/unit-10/3-electric-fields/study-guide/I5lSNgudkyVNrR1L)
- [10.4 Electric Potential Energy](/ap-physics-2-revised/unit-10/4-electric-potential-energy/study-guide/XB2Mvgn54KTHVQj6)
- [10.5 Electric Potential](/ap-physics-2-revised/unit-10/5-electric-potential/study-guide/NNaK6pYgyfxnT9Ma)
- [10.6 Capacitors](/ap-physics-2-revised/unit-10/6-capacitors/study-guide/BalONZFM1MOIxU84)
- [10.7 Conservation of Electric Energy](/ap-physics-2-revised/unit-10/7-conservation-of-electric-energy/study-guide/xeLRWRG6zabL3Yw4)

## Other Unit Tests

- [Unit 9: Thermodynamics](/ap-physics-2-revised/unit-test/9)
- [Unit 11: Electric Circuits](/ap-physics-2-revised/unit-test/11)
- [Unit 12: Magnetism and Electromagnetism](/ap-physics-2-revised/unit-test/12)
- [Unit 13: Geometric Optics](/ap-physics-2-revised/unit-test/13)
- [Unit 14: Waves, Sound, and Physical Optics](/ap-physics-2-revised/unit-test/14)
- [Unit 15: Modern Physics](/ap-physics-2-revised/unit-test/15)

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