AP Physics C: Electricity and Magnetism Practice Quiz: Gauss's Law
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) The magnitude of the electric field on the surface
B) The total charge enclosed by the surface
C) The surface area of the Gaussian surface
D) The total charge outside the surface
Correct Answer: B
Gauss's law, as stated by the equation $\Phi_{E}=\frac{q_{enc}}{\epsilon_{0}}$, directly relates the electric flux ($\Phi_{E}$) to the net charge enclosed (q_enc) by the surface.
A) A two-dimensional, open surface that follows the electric field lines.
B) A two-dimensional, closed loop in a uniform electric field.
C) A three-dimensional, closed surface used for applying Gauss's law.
D) A physical surface that contains an electric charge.
Correct Answer: C
The provided content explicitly states that 'A Gaussian surface is a three-dimensional, closed surface.' This is a fundamental requirement for the application of Gauss's law.
A) To maximize the amount of charge enclosed by the surface.
B) To ensure the electric field is zero everywhere on the surface.
C) To simplify the calculation of the surface integral $\oint\vec{E}\cdot d\vec{A}$.
D) To guarantee that the Gaussian surface is a perfect sphere or cylinder.
Correct Answer: C
The content states that Gaussian surfaces are constructed in this way 'resulting in a simplified surface integral.' When E is parallel to the surface normal (perpendicular to the surface), the dot product is maximal (E dA). When E is perpendicular to the surface normal (parallel to the surface), the dot product is zero. Both cases simplify the integral.
A) It is halved.
B) It remains the same.
C) It is doubled.
D) It is quadrupled.
Correct Answer: C
Based on the equation $\Phi_{E}=\frac{q_{enc}}{\epsilon_{0}}$, the electric flux is directly proportional to the enclosed charge, q_enc. Therefore, if the enclosed charge is doubled, the total electric flux is also doubled.
A) Zero
B) $\frac{+Q}{\epsilon_{0}}$
C) $\frac{-Q}{\epsilon_{0}}$
D) $\frac{+2Q}{\epsilon_{0}}$
Correct Answer: B
Gauss's law, $\Phi_{E}=\frac{q_{enc}}{\epsilon_{0}}$, only considers the charge enclosed (q_enc) by the surface. Since only the +Q charge is inside the surface, the charge -Q outside has no effect on the net flux. The net flux is therefore determined solely by the enclosed charge, +Q.
A) The total electric potential inside the surface.
B) The work done to move a charge around the surface.
C) The total electric flux through the closed surface.
D) The magnitude of the electric field at the surface.
Correct Answer: C
The equation $\oint\vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_{0}}$ is an expression of Gauss's law. The left side of the equation, the integral of the dot product of the electric field and the differential area vector over a closed surface, is the mathematical definition of total electric flux ($\Phi_{E}$).
A) The single point charge creating the electric field.
B) The net charge contained within the Gaussian surface.
C) The total charge in the entire system, both inside and outside the surface.
D) A small test charge placed on the surface.
Correct Answer: B
The subscript 'enc' on $q_{enc}$ stands for 'enclosed'. The content states that Gauss's law relates flux 'to the charge enclosed by that surface.' This means it is the algebraic sum (net charge) of all charges located inside the boundary of the Gaussian surface.
A) $\frac{+q}{\epsilon_{0}}$
B) $\frac{-q}{\epsilon_{0}}$
C) $\frac{+2q}{\epsilon_{0}}$
D) Zero
Correct Answer: D
Gauss's law depends on the net enclosed charge, $q_{enc}$. For an electric dipole, the total charge enclosed is $q_{enc} = (+q) + (-q) = 0$. Therefore, the net electric flux through the surface is $\Phi_{E}=\frac{0}{\epsilon_{0}} = 0$.
A) Determine the force between two point charges.
B) Calculate the electric potential of a charge distribution.
C) Describe the properties of a charge distribution.
D) Measure the value of the permittivity of free space, $\epsilon_{0}$.
Correct Answer: C
The first sentence of the provided content states that a key use is to 'Describe the properties of a charge distribution by applying Gauss's law.' While Gauss's law can be used to find the electric field (a property), this option is the most direct and encompassing answer based on the text.
A) The integral becomes zero.
B) The integral simplifies to E/A.
C) The integral simplifies to E × A.
D) The integral cannot be simplified without knowing the charge.
Correct Answer: C
When $\vec{E}$ is perpendicular to the surface, it is parallel to the area vector $d\vec{A}$. The dot product $\vec{E}\cdot d\vec{A}$ becomes simply E dA. Since E is also constant, it can be pulled out of the integral: $E \oint dA$. The integral of dA over the entire closed surface is the total surface area, A. Thus, the expression simplifies to E × A.