AP Physics C: Electricity and Magnetism, AP exams

Unit 8: Electric Charges, Fields, and Gauss's Law

Unit 8 builds electrostatics from charge through field, culminating in Gauss's law. The distinctive skill is computing the field of a continuous charge distribution, either by direct integration or by exploiting symmetry with a well-chosen Gaussian surface. Recognizing which method a given geometry calls for is itself a tested judgment.

Position: 1 of 6 units
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics C: Electricity and Magnetism. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Electric charge and charging processes: Conservation and quantization of charge, and charging by friction, conduction, and induction.
Coulomb's law and superposition: Force between point charges and the vector sum of contributions from multiple charges.
Electric fields: Field as force per unit charge, field-line representations, and fields of standard configurations.
Fields of continuous charge distributions: Integrating over lines, rings, disks, and shells to find the resulting field.
Electric flux: Flux through a surface, its dependence on orientation, and evaluating it for symmetric geometries.
Gauss's law: Relating enclosed charge to flux, and using spherical, cylindrical, and planar symmetry to find fields quickly.

Study tips

How to practice Charges, Fields, Gauss's Law.

Section-specific guidance that follows from how this material is usually tested.

Practice choosing the Gaussian surface first and justify the choice from the symmetry of the distribution.
Work at least one field-by-integration problem for a ring and for a rod so the setup is familiar.
Check limiting cases (far away, on axis, at the center) to validate any field expression you derive.

Where this fits

Unit 8: Electric Charges, Fields, and Gauss's Law is 1 of 6 units in the AP Physics C: Electricity and Magnetism guide. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics C: Electricity and Magnetism guide Official exam information

Sample Charges, Fields, Gauss's Law questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
Two point charges of 2.0 x 10^-6 C and 3.0 x 10^-6 C are held 0.30 m apart in vacuum. Using k = 9.0 x 10^9 N m^2/C^2, what is the magnitude of the electrostatic force between them?
1. A0.18 N
2. B0.20 N
3. C0.60 N
4. D1.8 N
Show answer
Answer: C. 0.60 N
Coulomb's law gives F = k(q1)(q2)/r^2 = (9.0 x 10^9)(6.0 x 10^-12)/(0.090) = 0.60 N. The 0.18 N choice divides by r instead of by r^2, and 1.8 N comes from writing (0.30)^2 as 0.030.
Key insight: Squaring the separation is the step most often skipped; 0.30 m gives 0.090 m^2, not 0.30.
Question 2Foundational
A point charge of 4.0 x 10^-6 C sits at the center of a spherical surface of radius 0.20 m. Using the permittivity of free space 8.85 x 10^-12 C^2/(N m^2), what is the total electric flux through that surface?
1. A3.5 x 10^-17 N m^2/C
2. B4.5 x 10^5 N m^2/C
3. C9.0 x 10^5 N m^2/C
4. D1.1 x 10^7 N m^2/C
Show answer
Answer: B. 4.5 x 10^5 N m^2/C
Gauss's law gives flux = q(enclosed)/epsilon0 = (4.0 x 10^-6)/(8.85 x 10^-12) = 4.5 x 10^5 N m^2/C. The 1.1 x 10^7 choice also divides by the radius squared, but the radius never appears because the flux depends only on the enclosed charge.
Key insight: Total flux through a closed surface counts only the charge inside, regardless of the surface's size or shape.
Question 3Intermediate
True or false: doubling the radius of a spherical Gaussian surface that encloses a single point charge at its center leaves the total electric flux through that surface unchanged.
1. ATrue
2. BFalse
Show answer
Answer: A. True
Gauss's law makes the flux equal to the enclosed charge divided by epsilon0, and doubling the radius does not change the enclosed charge. The field at the surface drops by a factor of 4 while the area grows by a factor of 4, so their product stays the same.
Key insight: Flux counts field lines crossing the surface, and no lines are created or destroyed between two radii in empty space.
Question 4Intermediate
A very long straight wire carries a uniform linear charge density of 2.0 x 10^-6 C/m. Using k = 9.0 x 10^9 N m^2/C^2, what is the magnitude of the electric field at a perpendicular distance of 0.50 m from the wire?
1. A1.8 x 10^4 N/C
2. B3.6 x 10^4 N/C
3. C7.2 x 10^4 N/C
4. D7.2 x 10^5 N/C
Show answer
Answer: C. 7.2 x 10^4 N/C
A cylindrical Gaussian surface gives E = 2k(lambda)/r = 2(9.0 x 10^9)(2.0 x 10^-6)/(0.50) = 7.2 x 10^4 N/C. The 3.6 x 10^4 N/C choice drops the factor of 2 that comes from the cylinder's lateral area.
Key insight: The field of an infinite line falls off as 1/r, not as 1/r^2, because the symmetry is cylindrical rather than spherical.
Question 5Advanced
A solid insulating sphere of radius R carries a total charge Q spread uniformly through its volume. In terms of k, Q, and R, what is the magnitude of the electric field at a point a distance R/2 from the center?
1. AZero
2. BkQ/(2R^2)
3. CkQ/R^2
4. D4kQ/R^2
Show answer
Answer: B. kQ/(2R^2)
A Gaussian sphere of radius R/2 encloses one eighth of the charge, so E = k(Q/8)/(R/2)^2 = kQ/(2R^2), which grows linearly with distance inside the sphere. The 4kQ/R^2 choice applies the external inverse-square result at r = R/2, and Zero would be correct only inside a conductor or a hollow shell.
Key insight: Inside a uniformly charged solid sphere the field is proportional to r; outside it falls off as 1/r^2.

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Unit 8: Electric Charges, Fields, and Gauss's Law