Gauss's Law Worked Examples Cheat Sheet

A printable reference covering electric flux, Gauss's law, symmetry, spherical, cylindrical, and planar charge distributions for grades 11-12.

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Gauss's law connects electric flux through a closed surface to the charge enclosed inside that surface. This cheat sheet helps students recognize when Gauss's law is the fastest way to find an electric field. It focuses on worked-example patterns for spherical, cylindrical, and planar symmetry. These patterns are important for AP Physics, honors physics, and early college electromagnetism.

Key Facts

Gauss's law states that the total electric flux through a closed surface is $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$ .
Electric flux through a flat surface in a uniform field is $\Phi_E = EA\cos\theta$ , where $\theta$ is the angle between $\vec{E}$ and the area vector.
For a point charge or spherical charge distribution, choose a spherical Gaussian surface so $E(4\pi r^2) = \frac{q_{\text{enc}}}{\varepsilon_0}$ .
Outside a spherically symmetric charge distribution, the field is $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$ , as if all charge were at the center.
For a long charged cylinder or line charge, choose a cylindrical Gaussian surface so $E(2\pi rL) = \frac{\lambda L}{\varepsilon_0}$ and $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ .
For an infinite plane sheet of charge, a pillbox Gaussian surface gives $2EA = \frac{\sigma A}{\varepsilon_0}$ and $E = \frac{\sigma}{2\varepsilon_0}$ .
For a conductor in electrostatic equilibrium, the electric field inside the conducting material is $E = 0$ and any excess charge lies on the surface.
Gauss's law is always true, but it is most useful when symmetry makes $\vec{E}$ constant in magnitude on the Gaussian surface or zero on parts of it.

Vocabulary

Electric flux: Electric flux measures how much electric field passes through a surface and is calculated by $\Phi_E = \int \vec{E} \cdot d\vec{A}$ .
Gaussian surface: A Gaussian surface is an imaginary closed surface chosen to make $\oint \vec{E} \cdot d\vec{A}$ easier to evaluate.
Enclosed charge: Enclosed charge is the net charge inside the Gaussian surface, written as $q_{\text{enc}}$ .
Permittivity of free space: The permittivity of free space is the constant $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ in Gauss's law.
Surface charge density: Surface charge density is charge per unit area, written as $\sigma = \frac{Q}{A}$ .
Line charge density: Line charge density is charge per unit length, written as $\lambda = \frac{Q}{L}$ .

Common Mistakes to Avoid

Using total charge instead of enclosed charge is wrong because Gauss's law uses only $q_{\text{enc}}$ inside the chosen closed surface.
Forgetting the dot product in $\Phi_E = \oint \vec{E} \cdot d\vec{A}$ is wrong because only the component of $\vec{E}$ perpendicular to the surface contributes to flux.
Choosing a Gaussian surface without symmetry is unhelpful because $E$ may not be constant, so $\oint \vec{E} \cdot d\vec{A}$ cannot be simplified easily.
Using $E = \frac{\sigma}{\varepsilon_0}$ for a single infinite sheet is wrong because one isolated sheet has $E = \frac{\sigma}{2\varepsilon_0}$ on each side.
Including flux through the flat ends of a cylindrical Gaussian surface for a line charge is wrong because $\vec{E}$ is parallel to those end surfaces, so $\vec{E} \cdot d\vec{A} = 0$ there.

Practice Questions

1 A point charge of $Q = 6.0 \times 10^{-9}\ \text{C}$ is at the center of a sphere with radius $r = 0.20\ \text{m}$ . What is the electric flux through the sphere?
2 An infinite line of charge has $\lambda = 3.0 \times 10^{-6}\ \text{C/m}$ . Find the electric field magnitude at $r = 0.050\ \text{m}$ from the line.
3 An infinite sheet has surface charge density $\sigma = 4.0 \times 10^{-8}\ \text{C/m}^2$ . What is the electric field magnitude on one side of the sheet?
4 A charge is outside a closed empty metal box. Explain why the net electric flux through the box is $0$ even though the electric field at points on the box may not be $0$ .

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