AP Physics C: E&M Unit 9 Practice Test

Pick a unit and take a timed, AP^®-style test built only from that unit's questions — with AI feedback on every free-response answer.

unit 8 unit 9 unit 10 unit 11 unit 12 unit 13

Electric Potential

Choose your test format

Two ways to practice unit 9

standard

The exam experience, sized to one unit — timed like the real thing

20 Multiple Choice Questions

40 min

1 FRQ – FRQ 1 – Mathematical Routines

24 min

1 FRQ – FRQ 2 – Translation Between Representations

24 min

Sections run back-to-back with the same directions, timers, and scoring as the real exam

Total: 1 hr 28 min

customizable

Pick your own mix of question types

multiple choice questions

free response questions (choose as many as you want)

FRQ 1 – Mathematical Routines · 24 minFRQ 2 – Translation Between Representations · 24 minFRQ 3 – Experimental Design · 24 minFRQ 4 – Qualitative/Quantitative Translation · 24 min

Total: 40 min

How scoring works: multiple choice is scored instantly, free-response answers get point-by-point AI feedback, and your result is a percentage for this unit — not a projected AP^® score.

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AP Physics C: E&M Unit 9 practice questions

Try 10 multiple-choice questions from Unit 9 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

Question 1
A point charge $+Q$ creates an electric potential $V_0$ at a distance $R$ . If the charge is changed to $-2Q$ and the distance is reduced to $R/2$ , what is the new electric potential at that point?
$-4V_0$
$-2V_0$
$+2V_0$
$+4V_0$
Show answer
Answer: (A)
Applying the functional dependence $V = kQ/r$ requires substituting the new values. Substituting $-2Q$ for charge and $R/2$ for distance yields $V_{new} = k(-2Q)/(R/2) = -4(kQ/R) = -4V_0$ .
Question 2
A uniformly charged ring of radius $R$ and total charge $Q$ produces an electric potential $V_0$ at a point on its axis at distance $x = R$ from the center. If the ring's radius and the distance to the point are both doubled to $2R$ , what is the new potential?
$V_0/4$
$V_0/2$
$V_0$
$2V_0$
Show answer
Answer: (B)
Using the potential formula for a ring $V = kQ/\sqrt{R^2 + x^2}$ , doubling both $R$ and $x$ results in a denominator of $\sqrt{(2R)^2 + (2x)^2} = 2\sqrt{R^2 + x^2}$ , reducing the potential by a factor of 2.
Question 3
A thin semicircular rod with total charge $Q$ and radius $R$ produces an electric potential $V_c$ at its center of curvature. If the radius is doubled to $2R$ while keeping the total charge $Q$ constant, what is the new potential at the center?
$V_c/4$
$V_c/2$
$V_c$
$2V_c$
Show answer
Answer: (B)
Since every element of charge $dq$ is at the same distance $R$ , the potential is $V = \int k dq/R = kQ/R$ . Doubling $R$ while keeping $Q$ constant halves the potential.
Question 4
An infinitely long wire with linear charge density $\lambda$ creates a potential difference $\Delta V$ between points at distances $r$ and $2r$ from the wire. If the linear charge density is doubled to $2\lambda$ , what is the new potential difference between the same points?
$\Delta V/2$
$\Delta V$
$2\Delta V$
$4\Delta V$
Show answer
Answer: (C)
The potential difference near an infinite wire is $\Delta V = (\lambda/2\pi\epsilon_0)\ln(r_2/r_1)$ . Since $\Delta V$ is directly proportional to $\lambda$ , doubling $\lambda$ doubles the potential difference.
Question 5
Three point charges, each with charge $+q$ , are fixed at the vertices of an equilateral triangle of side length $d$ . The total electric potential energy of the system is $U_0$ . If the side length of the triangle is increased to $3d$ , what is the new potential energy?
$U_0/9$
$U_0/3$
$3U_0$
$9U_0$
Show answer
Answer: (B)
The potential energy of the system is the sum of pairwise interactions $U = \sum k q_i q_j / r_{ij}$ . Since every distance $r_{ij}$ scales by 3, the total energy scales by $1/3$ .
Question 6
A finite rod of length $L$ has a uniform linear charge density $\lambda$ . The electric potential at a point on the axis at distance $d$ from one end is $V_0$ . If the linear charge density is doubled while $L$ and $d$ remain constant, what is the new potential?
$V_0/4$
$V_0/2$
$2V_0$
$4V_0$
Show answer
Answer: (C)
Electric potential is directly proportional to the charge density $\lambda$ (since $V = \int k \lambda dx / r$ ). Doubling $\lambda$ directly doubles the potential.
Question 7
A conducting sphere of radius $R$ carries a net charge $Q$ , resulting in an electric potential $V_s$ at its surface. If the sphere's radius is reduced to $R/3$ while the charge $Q$ remains constant, what is the new potential at the surface?
$V_s/9$
$V_s/3$
$3V_s$
$9V_s$
Show answer
Answer: (C)
The potential at the surface of a conducting sphere is $V = kQ/R$ . Reducing $R$ by a factor of 3 increases the potential by a factor of 3.
Question 8
Two positive point charges $+Q$ are separated by a distance $2x$ . The electric potential at the midpoint between them is $V_0$ . If the distance between the charges is increased to $4x$ , what is the new potential at the midpoint?
$V_0/4$
$V_0/2$
$2V_0$
$4V_0$
Show answer
Answer: (B)
The potential at the midpoint is $V = 2(kQ/x)$ . If the separation doubles to $4x$ , the distance from midpoint to charge becomes $2x$ , reducing $V$ to $2(kQ/2x) = V_0/2$ .
Question 9
A circular arc of radius $R$ subtends an angle $\theta$ at its center and carries a uniform linear charge density $\lambda$ . The potential at the center is $V_0$ . If the angle is increased to $2\theta$ while $R$ and $\lambda$ remain constant, what is the new potential?
$V_0/2$
$V_0$
$2V_0$
$4V_0$
Show answer
Answer: (C)
The potential is $V = \int k dq/R = k(R\theta\lambda)/R = k\lambda\theta$ . Doubling the angle $\theta$ doubles the total charge and thus doubles the potential.
Question 10
A spherical shell of radius $R$ carries charge $+Q$ . A second concentric shell of radius $2R$ carries charge $+Q$ . The potential at the center is $V_{center}$ . If the outer shell is removed, what is the new potential at the center?
$1/3 V_{center}$
$2/3 V_{center}$
$3/4 V_{center}$
$4/3 V_{center}$
Show answer
Answer: (B)
Initial potential is $V = kQ/R + kQ/2R = 1.5 kQ/R$ . Removing the outer shell leaves $V = kQ/R$ . The ratio is $1/1.5 = 2/3$ .

What's on the AP Physics C: E&M Unit 9 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

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

Questions are drawn from 530 multiple-choice questions and 39 FRQs written for Unit 9, Electric Potential.

Practice AP Physics C: E&M Unit 9 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

FRQ 1 – Mathematical Routines (11 questions)
FRQ 2 – Translation Between Representations (3 questions)
FRQ 3 – Experimental Design (5 questions)
FRQ 4 – Qualitative/Quantitative Translation (20 questions)

AP Physics C: E&M Unit 9 Practice Test

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AP Physics C: E&M Unit 9 practice questions

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AP Physics C: E&M Unit 9 Practice Test

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