An infinite sheet has surface charge density σ = 4.010^-6 C per m squared. Find the field magnitude it produces (_0 = 8.8510^-12). [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does Gauss's law use symmetry to find the electric field of a charge distribution?

Topic 8.6 Gauss's Law: apply Gauss's law with a chosen Gaussian surface to find the field of spherically, cylindrically and planar-symmetric charge distributions.

A calculus-based answer to AP Physics C E&M Topic 8.6, covering Gauss's law, choosing a Gaussian surface, and deriving the field of spheres, lines and planes, plus the field inside conductors.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Gauss's law states that the net electric flux through any closed surface equals the enclosed charge over $\varepsilon_0$ : $\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$ . It is always true, but it lets you solve for the field only when symmetry makes $E$ constant on a chosen Gaussian surface and the field either perpendicular (full flux) or parallel (zero flux) to each part. The three symmetric cases give standard fields: a sphere or point charge, $E = \dfrac{kQ}{r^2}$ outside; an infinite line or cylinder, $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ ; an infinite plane of charge, $E = \dfrac{\sigma}{2\varepsilon_0}$ (uniform). Inside a solid uniformly charged sphere the field grows linearly, $E = \dfrac{kQr}{R^3}$ , and inside a conductor it is zero.

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What this topic is asking
Gauss's law
Choosing a Gaussian surface
The conductor result
Try this

What this topic is asking

The College Board (Topic 8.6) wants you to apply Gauss's law to find the electric field of charge distributions with spherical, cylindrical or planar symmetry, by choosing a Gaussian surface on which the field is constant and either perpendicular or parallel to the surface. This is the most powerful tool in electrostatics: where the symmetry is right, a hard integral collapses to a one-line calculation.

Gauss's law

The law is always valid, but it is only useful for finding the field when the charge distribution has enough symmetry that you can pull $E$ out of the integral. Otherwise you fall back to the direct integration of Topic 8.4.

Choosing a Gaussian surface

The art is picking a closed surface, the Gaussian surface, that matches the symmetry so that on each part either:

the field has constant magnitude and is perpendicular to the surface (so $\vec{E}\cdot d\vec{A} = E\,dA$ and the flux is $E\times\text{area}$ ), or
the field is parallel to the surface (so $\vec{E}\cdot d\vec{A} = 0$ and that part contributes nothing).

This turns the surface integral into $E\times(\text{area with flux}) = q_{enc}/\varepsilon_0$ , which you solve for $E$ .

Symmetry	Gaussian surface	Field result
Spherical (point, sphere, shell)	Concentric sphere	$E = \dfrac{kQ}{r^2}$ outside
Cylindrical (line, long cylinder)	Coaxial cylinder	$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
Planar (infinite sheet)	Pillbox through the sheet	$E = \dfrac{\sigma}{2\varepsilon_0}$

The conductor result

Field of a uniformly charged solid sphere

A solid insulating sphere of radius $R$ carries charge $Q$ spread uniformly through its volume. Find the field magnitude both outside ( $r \ge R$ ) and inside ( $r < R$ ).

step 1 Choose the Gaussian surface

Spherical symmetry means a concentric Gaussian sphere of radius $r$ . On it, $\vec{E}$ is radial and constant in magnitude, so $\oint\vec{E}\cdot d\vec{A} = E(4\pi r^2)$ .

step 2 Outside the sphere ( $r \ge R$ )

The whole charge is enclosed: $q_{enc} = Q$ . Gauss: $E(4\pi r^2) = \dfrac{Q}{\varepsilon_0}$ , so $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{kQ}{r^2}$ . The sphere acts like a point charge at its center.

step 3 Find the enclosed charge inside ( $r < R$ )

Uniform volume density $\rho = \dfrac{Q}{\tfrac{4}{3}\pi R^3}$ . The Gaussian sphere encloses $q_{enc} = \rho\cdot\tfrac{4}{3}\pi r^3 = Q\dfrac{r^3}{R^3}$ .

step 4 Apply Gauss's law inside

$E(4\pi r^2) = \dfrac{q_{enc}}{\varepsilon_0} = \dfrac{Q r^3/R^3}{\varepsilon_0}$ , so $E = \dfrac{Q r}{4\pi\varepsilon_0 R^3} = \dfrac{kQr}{R^3}$ . The field grows linearly from zero at the center to $\dfrac{kQ}{R^2}$ at the surface, matching the outside result there.

Try this

Q1. State the field inside a hollow conducting sphere that carries net charge but no charge in its cavity. [1 point]

Cue. Zero: a Gaussian surface inside the conductor encloses no charge.

Q2. An infinite sheet has surface charge density $\sigma = 4.0\times10^{-6}$ C per m squared. Find the field magnitude it produces ( $\varepsilon_0 = 8.85\times10^{-12}$ ). [2 points]

Cue. $E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{4.0\times10^{-6}}{2(8.85\times10^{-12})} = 2.3\times10^5$ N/C, uniform.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). A solid insulating sphere of radius

R

has charge

Q

spread uniformly through its volume. At a point inside, a distance

r < R

from the center, the field magnitude is (A)

\dfrac{kQ}{r^2}

(B)

\dfrac{kQr}{R^3}

(C)

\dfrac{kQ}{R^2}

(D) zero. Justify your reasoning.

Show worked answer →

A 1-point MCQ on Gauss's law inside a uniform sphere. The answer is (B).

A Gaussian sphere of radius $r$ encloses only the fraction of charge inside it: $q_{enc} = Q\dfrac{r^3}{R^3}$ . Gauss's law gives $E(4\pi r^2) = \dfrac{q_{enc}}{\varepsilon_0}$ , so $E = \dfrac{q_{enc}}{4\pi\varepsilon_0 r^2} = \dfrac{kQr^3/R^3}{r^2} = \dfrac{kQr}{R^3}$ , which grows linearly with $r$ . The trap is (A), the point-charge result, which only holds outside ( $r \ge R$ ).

AP 2022 (style)6 marksSection II (FRQ, derivation). An infinitely long straight wire carries a uniform linear charge density

\lambda

. (a) Describe the Gaussian surface you choose and why. (b) Apply Gauss's law to derive the field magnitude a distance

r

from the wire. (c) State how the field depends on

r

and contrast this with a point charge.

Show worked answer →

A 6-point FRQ deriving the line-charge field with Gauss's law.

(a) Surface (2 points): a coaxial cylinder of radius $r$ and length $L$ . By symmetry the field is radial and constant in magnitude over the curved side, and the field is parallel to the flat end caps (zero flux there).
(b) Derivation (3 points): flux $= \oint\vec{E}\cdot d\vec{A} = E(2\pi r L)$ (only the curved side contributes). Enclosed charge $= \lambda L$ . Gauss: $E(2\pi r L) = \dfrac{\lambda L}{\varepsilon_0}$ , so $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ .
(c) Dependence (1 point): $E \propto 1/r$ , falling off more slowly than a point charge's $1/r^2$ , because of the line geometry.

Markers reward the cylinder with zero-flux caps, the $E(2\pi rL) = \lambda L/\varepsilon_0$ step, and the $1/r$ result.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)