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AP Physics C: Electricity and Magnetism Practice Quiz: Electric Fields of Charge Distributions

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

The equation $\vec{E}=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}\hat{r}$ is used to calculate the electric field of a continuous charge distribution. This equation is a direct mathematical representation of which physical principle?

All Questions (9)

The equation $\vec{E}=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}\hat{r}$ is used to calculate the electric field of a continuous charge distribution. This equation is a direct mathematical representation of which physical principle?

A) Gauss's Law

B) The principle of superposition

C) Conservation of charge

D) Ampere's Law

Correct Answer: B

The integral sums the infinitesimal electric field contributions (dE) from each infinitesimal charge element (dq) over the entire distribution. This process of vectorially adding up the fields from individual parts to find the total field is the principle of superposition.

In setting up the integral to find the electric field of a uniformly charged thin rod of length L and total charge Q, what is the correct expression for the infinitesimal charge element dq in terms of an infinitesimal length element dx?

A) dq = Q dx

B) dq = (L/Q) dx

C) dq = (Q/L) dx

D) dq = QL dx

Correct Answer: C

For a uniformly charged object, the linear charge density (λ) is the total charge divided by the total length, so λ = Q/L. The infinitesimal charge dq in a small length dx is given by dq = λ dx, which is equivalent to dq = (Q/L) dx.

A thin, circular ring of radius R is located in the xy-plane and centered at the origin. It carries a uniform positive charge. Based on symmetry considerations, what is the direction of the net electric field at a point P located on the positive z-axis?

A) The field is zero.

B) In the positive z-direction, away from the ring.

C) In the negative z-direction, toward the ring.

D) Radially outward in the xy-plane.

Correct Answer: B

Due to the ring's symmetry, for every charge element dq on the ring creating an electric field component in the xy-plane, there is a diametrically opposite charge element creating a component that cancels it. Therefore, all xy-components cancel out, and only the z-components, which all point in the same direction (away from the positive ring), add up.

Consider a uniformly charged, thin rod of length 2L centered at the origin and lying along the x-axis. Which statement best explains why the x-component of the electric field is zero at a point P on the y-axis?

A) The y-components of the field from symmetric charge elements cancel out.

B) The x-components of the field from symmetric charge elements on the left and right of the origin are equal and opposite, thus they cancel out.

C) The distance from point P to any charge element on the rod is constant.

D) The total charge on the rod is zero.

Correct Answer: B

For any charge element dq at a position +x on the rod, there is an identical charge element at -x. At a point on the y-axis, the electric field vector from +x has a negative x-component, while the vector from -x has a positive x-component of the same magnitude. These two x-components cancel each other. This is true for all such symmetric pairs, so the total x-component of the field is zero.

When calculating the electric field at a point P due to a charged arc, the term 'r' in the integral $\vec{E}=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}\hat{r}$ represents which quantity?

A) The radius of the arc.

B) The distance from the center of the arc to point P.

C) The distance from an infinitesimal charge element dq to point P.

D) The length of the arc.

Correct Answer: C

The integral sums the contributions of all infinitesimal charge elements dq. The term 'r' in the integrand represents the distance from the specific charge element dq being considered to the point P where the field is being calculated. This distance may or may not be constant, depending on the geometry of the distribution and the location of P.

A solid, uniformly charged sphere and a thin, uniformly charged spherical shell have the same total charge Q and the same radius R. How do their electric fields compare at a distance r = 2R from their centers?

A) The field of the solid sphere is stronger.

B) The field of the spherical shell is stronger.

C) The fields are equal.

D) The fields are both zero.

Correct Answer: C

Due to the spherical symmetry of both charge distributions, for any point outside the sphere (r > R), the electric field is identical to that of a point charge Q located at the center. Since both distributions have the same total charge Q, their electric fields at r = 2R will be identical.

Which of the following best describes the general procedure for finding the electric field of a continuous charge distribution using integration?

A) Calculate the total charge and divide by the area of a surrounding surface.

B) Assume the entire charge is a point charge at the geometric center.

C) Divide the distribution into infinitesimal charges, calculate the field from each, and sum (integrate) these contributions as vectors.

D) Find the electric potential and multiply by the total charge.

Correct Answer: C

The method of integration for electric fields is an application of the principle of superposition. The continuous distribution is broken down into an infinite number of infinitesimal point charges (dq). The electric field (dE) from each dq is calculated using Coulomb's Law, and then all these vector fields are summed via integration to find the net electric field.

A square loop of wire with side length s lies in the xy-plane, centered at the origin. If the loop has a uniform positive charge, what can be concluded about the electric field at a point (0, 0, z) on the z-axis due to symmetry?

A) The electric field has only x and y components.

B) The electric field has only a z-component.

C) The electric field is zero everywhere on the z-axis.

D) Symmetry considerations are not sufficient to simplify the analysis.

Correct Answer: B

The analysis is similar to that of a charged ring. For any charge element dq on one side of the square, there is a corresponding element on the opposite side. The components of their electric fields parallel to the xy-plane will point in opposite directions and cancel out. Only the components along the z-axis will add constructively. Therefore, the net electric field must point along the z-axis.

In the integral expression for the electric field, $\vec{E}=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}\hat{r}$, the unit vector $\hat{r}$ points in which direction?

A) From the origin to the point where the field is being calculated.

B) From the origin to the infinitesimal charge element dq.

C) From the infinitesimal charge element dq to the point where the field is being calculated.

D) In the direction of the total electric field.

Correct Answer: C

The term inside the integral represents the electric field contribution, dE, from a single infinitesimal charge element, dq. This field is a vector that points from the source of the field (the charge dq) to the point of interest (where the field is being calculated). Therefore, the unit vector r-hat must point in this direction.