Unit 8 · Topic 8.6

8.6 Gauss’s Law

Gauss's law says the net electric flux through any closed surface equals the charge inside divided by ε₀. With spherical, cylindrical or planar symmetry, it turns a hard field calculation into a few lines of algebra. It's also the first of Maxwell's four equations of electromagnetism.

Key terms

Gauss’s law
Gaussian surface
enclosed charge
spherical, cylindrical and planar symmetry
Maxwell’s equations

The law

$\oint\vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$

The closed surface you choose is called a Gaussian surface. It's imaginary: you pick it to make the math easy. $q_{\text{enc}}$ is the net charge inside it. Charges outside the surface make fields that pass through it, but they contribute zero net flux.

The law is always true, but it only gives you E when symmetry lets you pull E out of the integral. That happens when, on each part of the surface, the field is either perpendicular to the surface with the same size everywhere, or parallel to it (zero flux).

Why it works

In 8.5 you saw that the flux of a point charge through a sphere around it is q/ε₀, whatever the sphere's radius. Squash or stretch the sphere into any closed shape and the same field lines still cross it, so the flux is unchanged. A charge outside the surface sends every line in one side and out the other, adding zero. Superposition then extends this to any collection of charges, which is Gauss's law.

This only works because the field falls off exactly as 1/r². For charges at rest, Gauss's law and Coulomb's law are two ways of stating the same physics. Gauss's law also holds when charges move, which is why it's the first of Maxwell's four equations, the set that describes all of electromagnetism.

Choosing the Gaussian surface

Symmetry	Gaussian surface	Result
Sphere or point charge	Concentric sphere of radius r	$E(4\pi r^2) = q_{\text{enc}}/\varepsilon_0$
Long line or cylinder	Coaxial cylinder of radius r, length ℓ	$E(2\pi r\ell) = q_{\text{enc}}/\varepsilon_0$
Large flat sheet	Short cylinder (pillbox) poking through both sides	$E(2A) = \sigma A/\varepsilon_0$

Key results

Outside any spherically symmetric charge Q: $E = \dfrac{kQ}{r^2}$ , as if all the charge were at the center.
Inside a uniformly charged insulating sphere of radius R: $q_{\text{enc}} = Q\dfrac{r^3}{R^3}$ , so $E = \dfrac{kQr}{R^3}$ . The field grows linearly from zero at the center to its peak at the surface.
Inside a uniformly charged spherical shell with nothing in its hollow, and inside the material of any conductor in equilibrium: E = 0. For the shell, $q_{\text{enc}} = 0$ and symmetry together give this; zero enclosed charge alone isn't enough.
Infinite line or outside a long cylinder: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ .
Infinite sheet of charge: $E = \dfrac{\sigma}{2\varepsilon_0}$ , the same at every distance.

When the density isn't uniform

If ρ depends on r, you can't multiply ρ by a volume. Integrate instead, using thin shells: for a sphere, $q_{\text{enc}} = \displaystyle\int_0^r\rho(r')\,4\pi r'^2\,dr'$ . For a cylinder of length ℓ, use shells of volume $2\pi r'\ell\,dr'$ . Then put that charge into Gauss's law.

Graphs of E against r

For a uniformly charged solid sphere, the graph of E against r is a straight line rising from the origin to r = R, then a 1/r² curve falling off outside. It's continuous at the surface. For a charged conducting sphere, E is zero up to R, jumps to kQ/R² at the surface, then falls as 1/r². Being able to sketch and explain these graphs is a core skill.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Uniformly charged insulating sphere
An insulating sphere of radius 0.10 m carries +6.0 nC spread uniformly through its volume. Find the field at r = 0.050 m and at r = 0.20 m from the center.
Show the solutionHide the solution
1. Step 1: Inside (r = 0.050 m): a Gaussian sphere encloses $Q(r/R)^3$ . Gauss's law gives $E = \dfrac{kQr}{R^3} = \dfrac{(9.0\times10^{9})(6.0\times10^{-9})(0.050)}{(0.10)^3} = 2.7\times10^{3}$ N/C.
2. Step 2: Outside (r = 0.20 m): the Gaussian sphere encloses all of Q, so $E = \dfrac{kQ}{r^2} = \dfrac{(9.0\times10^{9})(6.0\times10^{-9})}{(0.20)^2} = 1.35\times10^{3}$ N/C.
3. Step 3: Both fields point radially outward.
Answer: 2.7 × 10³ N/C at 0.050 m; 1.35 × 10³ N/C at 0.20 m; both outward
Example 2Calculator allowed
Field of a long line of charge
Use Gauss's law to find the field a distance r from a very long, thin line with uniform charge density λ.
Show the solutionHide the solution
1. Step 1: By symmetry, the field points straight out from the line and has the same size everywhere at distance r.
2. Step 2: Choose a cylinder of radius r and length ℓ around the line. On the curved side, $\vec{E}$ is parallel to $d\vec{A}$ with constant size, so its flux is $E(2\pi r\ell)$ . On the flat ends, $\vec{E}$ runs along the surface, so their flux is zero.
3. Step 3: Enclosed charge: $\lambda\ell$ . Gauss's law: $E(2\pi r\ell) = \dfrac{\lambda\ell}{\varepsilon_0}$ , so $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ .
Answer: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ , directed radially away from the line (for positive λ)
Example 3Calculator allowed
Nonuniform density (classic trap)
A sphere of radius R has charge density $\rho = Ar$ , where A is a positive constant. Find the field inside the sphere at distance r from the center.
Show the solutionHide the solution
1. Step 1: The density changes with r, so $\rho \cdot \tfrac{4}{3}\pi r^3$ is wrong. That's the trap.
2. Step 2: Add thin shells: $q_{\text{enc}} = \displaystyle\int_0^r Ar'\,(4\pi r'^2)\,dr' = 4\pi A\frac{r^4}{4} = \pi A r^4$ .
3. Step 3: Gauss's law: $E(4\pi r^2) = \dfrac{\pi A r^4}{\varepsilon_0}$ , so $E = \dfrac{Ar^2}{4\varepsilon_0}$ , pointing outward.
Answer: $E = \dfrac{Ar^2}{4\varepsilon_0}$ , radially outward

Common mistakes

Using the total charge instead of the charge enclosed by the Gaussian surface when the point is inside the distribution.
Thinking Gauss's law fails without symmetry. It's always true; it's just not useful for finding E without symmetry.
Concluding that E = 0 everywhere on a surface because the net flux is zero. Zero net flux only means zero enclosed charge.
Using the conductor result σ/ε₀ for a lone sheet of charge. An isolated sheet gives σ/(2ε₀).

On the exam

Free-response questions often ask you to state the Gaussian surface, justify the symmetry in words, and then apply the law. Each of those steps can earn a point.
Be ready to sketch E against r across several regions (inside, in a shell, outside) and explain any jumps or zeros.

Connected topics

Videos

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