AP Physics C: E&M — Practice Questions flashcards

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Multiple-choice practice questions, one best answer each, on Gauss's law, potential, conductors, capacitors and dielectrics, RC circuits, Biot–Savart and Ampère's law, induction, and RL and LC circuits.

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Practice deck · part of AP Physics C: Electricity and Magnetism · written against: College Board AP Physics C: Electricity and Magnetism Course and Exam Description (2024 revision). Updated 2026-10-10.

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A charge $+q$ is fixed at $x = 0$ and a charge $-4q$ is fixed at $x = d$ ⁠. At what finite point on the x-axis is the net electric field zero?

A) $x = 2d$
B) $x = -d$
C) $x = -\frac{d}{3}$
D) $x = \frac{d}{3}$

B) $x = -d$

For opposite charges the fields can cancel only outside the pair, on the side of the smaller charge: $\frac{kq}{r^2} = \frac{4kq}{\left(r + d\right)^2}$ gives $r = d$ ⁠. The point $\frac{d}{3}$ would be right if both charges had the same sign, but between opposite charges the two fields point the same way.
Electric Charges Fields and Gauss's Law › Coulomb's Law and Fields
In a region of space the electric potential is $V = 2x^2y - 3y$ ⁠, with $V$ in volts and $x$ and $y$ in meters. What is the electric field at the point $\left(1, 2\right)$ ⁠?

A) $E_x = -8$ V/m and $E_y = -1$ V/m
B) $E_x = -8$ V/m and $E_y = +1$ V/m
C) $E_x = +8$ V/m and $E_y = -1$ V/m
D) $E_x = -4$ V/m and $E_y = +1$ V/m

B) $E_x = -8$ V/m and $E_y = +1$ V/m

Each component is minus a partial derivative: $E_x = -4xy = -8$ and $E_y = -\left(2x^2 - 3\right) = +1$ at $\left(1, 2\right)$ ⁠. Leaving out the minus sign flips both components.
Electric Potential
A point charge $+q$ sits at the center of a hollow conducting spherical shell that carries a net charge of $-3q$ ⁠. In electrostatic equilibrium, what charges are on the shell's inner and outer surfaces?

A) Inner $0$ and outer $-3q$
B) Inner $+q$ and outer $-4q$
C) Inner $-q$ and outer $-2q$
D) Inner $-q$ and outer $-3q$

C) Inner $-q$ and outer $-2q$

For zero field inside the metal, the inner surface must hold $-q$ to cancel the central charge, and the remaining $-3q - \left(-q\right) = -2q$ goes to the outer surface. Putting $-3q$ on the outer surface would give the shell a total of $-4q$ ⁠.
Conductors and Capacitors › Conductors
In a wire of radius $R$ ⁠, the current density falls off from the axis as $J = J_0\left(1 - \frac{r}{R}\right)$ ⁠. What is the total current in the wire?

A) $\frac{\pi J_0R^2}{3}$
B) $\frac{2\pi J_0R^2}{3}$
C) $\frac{\pi J_0R^2}{2}$
D) $\pi J_0R^2$

A) $\frac{\pi J_0R^2}{3}$

Integrating over rings of area $2\pi r dr$ ⁠: $\int_0^R J_0\left(1 - \frac{r}{R}\right)2\pi r dr = 2\pi J_0\left(\frac{R^2}{2} - \frac{R^2}{3}\right)$ ⁠. Multiplying the average of $J$ along a radius by the area gives $\frac{\pi J_0R^2}{2}$ ⁠, which wrongly weights the center as much as the edge.
Electric Circuits › Current Resistance and Power
A charged particle moves in a circle perpendicular to a uniform magnetic field. If its speed is doubled, what happens to the radius and the period of its circular motion?

A) The radius doubles and the period doubles.
B) The radius doubles and the period is unchanged.
C) The radius is unchanged and the period halves.
D) The radius quadruples and the period is unchanged.

B) The radius doubles and the period is unchanged.

$qvB = \frac{mv^2}{r}$ gives $r = \frac{mv}{qB}$ ⁠, proportional to speed, while $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$ does not depend on speed. A faster particle covers a larger circle in the same time.
Magnetic Fields and Electromagnetism › Magnetic Forces
A rectangular loop has its sides of length $a$ parallel to a long wire carrying steady current $I$ ⁠. The loop moves directly away from the wire at speed $v$ ⁠; its near side is at distance $r$ and its far side at $r + b$ ⁠. What is the emf?

A) zero, because the current is steady
B) $\frac{\mu_0Iav}{2\pi}\left(\frac{1}{r} - \frac{1}{r + b}\right)$
C) $\frac{\mu_0Iav}{2\pi r}$
D) $\frac{\mu_0Iav}{2\pi}\ln\left(\frac{r + b}{r}\right)$

B) $\frac{\mu_0Iav}{2\pi}\left(\frac{1}{r} - \frac{1}{r + b}\right)$

Each side parallel to the wire has motional emf $Bav$ ⁠, and the two oppose each other, leaving $av\left(B\left(r\right) - B\left(r + b\right)\right)$ ⁠; differentiating the flux $\frac{\mu_0Ia}{2\pi}\ln\left(\frac{r + b}{r}\right)$ gives the same. A steady current still gives an emf, because the flux through the moving loop changes.
Electromagnetic Induction › Faraday's and Lenz's Laws

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Topics follow the College Board AP Physics C: Electricity and Magnetism Course and Exam Description (2024 revised framework, Units 8–13); the AP name is used nominatively and no course or exam text is reproduced.
OpenStax University Physics Volume 2, consulted as a reference; no text copied.
Formulas and results verified with symbolic and numerical calculation.

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