Gauss's Law

Gauss's law simply says that the total field poking out of any closed bag tells you how much charge is inside the bag. It's named after Carl Friedrich Gauss, the mathematician who worked it out in the 1830s. It's good for two things: checking how much charge a region holds, and, when the charge is arranged symmetrically, finding the field with almost no calculation. The formula $\oint_S \mathbf{E}\cdot d\mathbf{A} = Q_{\text{enc}}/\varepsilon_0$ looks intimidating, but it's just $\oint_S$ (add up over the whole closed surface), $\mathbf{E}\cdot d\mathbf{A}$ (how much field goes out through each tiny patch) and $Q_{\text{enc}}/\varepsilon_0$ (the charge inside, divided by a fixed constant). Field coming out on the left, charge inside on the right; bear with it.

Field lines as bookkeeping. The last lesson found that a point charge pushes exactly $q/\varepsilon_0$ of flux through any sphere around it. Gauss's law says much more: through any closed surface, the net outward flux is $Q_{\text{enc}}/\varepsilon_0$. Think of each charge as sprouting a fixed number of field lines. Wrap any bag around some charges, count the lines leaving minus the lines entering, and you've counted the charge inside.

Why it's true, for any shape.

• A point charge inside. A small patch $dA$ at distance $r$, tilted by $\alpha$ from facing the charge, catches flux $\frac{kq}{r^2}\cos\alpha\,dA$. But $\frac{\cos\alpha\,dA}{r^2}$ is the patch's solid angle $d\Omega$: its size as seen from the charge, the area of its shadow on a sphere of radius 1. So the patch's flux is $kq\,d\Omega$, and a closed surface wraps all the way around, through a total solid angle of $4\pi$, giving $4\pi kq = q/\varepsilon_0$ whatever its shape.
• A point charge outside. Each thin cone of directions from the charge pierces the surface twice, once going in and once coming out, with equal and opposite flux. Net: zero.
• Many charges. Fields add, so fluxes add: $\Phi = \sum q_i/\varepsilon_0$ over the charges inside. Continuous charge is the limit of many small ones.

The law holds for every closed surface, but it only gives you one number, the total flux. So what can we do with that? It finds $E$ only when symmetry lets you pull $E$ out of the integral.

The recipe.

1. From the symmetry, decide which way $\mathbf{E}$ points and on which surfaces its strength is constant.
2. Choose a closed Gaussian surface built from pieces where $\mathbf{E}$ is either perpendicular to the surface with constant strength ($\mathbf{E}\cdot d\mathbf{A} = E\,dA$) or parallel to it ($\mathbf{E}\cdot d\mathbf{A} = 0$).
3. Find $Q_{\text{enc}}$, set the flux equal to $Q_{\text{enc}}/\varepsilon_0$ and solve for $E$.

Sphere. For spherically symmetric charge, use a concentric sphere of radius $r$: $E\cdot4\pi r^2 = Q_{\text{enc}}/\varepsilon_0$. Outside the charge, $E = \frac{Q}{4\pi\varepsilon_0r^2}$: any spherical ball or shell acts like a point charge at its center (the shell theorem, true of gravity for the same reason). Inside a uniform ball of radius $R$, only $Qr^3/R^3$ is enclosed, so $E = \frac{kQr}{R^3}$ grows linearly from zero. Inside a hollow shell nothing is enclosed, and $E = 0$.

Line. For a long straight line with $\lambda$ coulombs per meter, use a coaxial can of radius $r$ and length $L$. The field is radial, so the end caps catch nothing and the side catches $E\cdot2\pi rL$. That equals $\lambda L/\varepsilon_0$, so $E = \frac{\lambda}{2\pi\varepsilon_0r} = \frac{2k\lambda}{r}$: the result that took a trig substitution in Fields of Continuous Charge.

Plane. For a large flat sheet with $\sigma$ coulombs per square meter, use a pillbox straddling the sheet. Field lines leave through both faces, so $2EA = \sigma A/\varepsilon_0$ and $E = \frac{\sigma}{2\varepsilon_0}$ at every distance: the lines never spread out, so the field never weakens. Sheets carrying $+\sigma$ and $-\sigma$ side by side add to $\sigma/\varepsilon_0$ between them and cancel outside, which is the field inside a capacitor, coming up in the next unit.

A common mistake. It's tempting to think the $\mathbf{E}$ in the integral is only the field of the charges inside, since those are the only ones that count. It isn't: $\mathbf{E}$ is the total field, outside charges included. Their part just adds zero net flux, because each of their field lines enters and leaves. Two other slips: counting charge that sits outside the surface, and trying the law on a short rod or a small disk. The law is still true there, but $E$ isn't constant on any convenient surface, so it can't come out of the integral.

High-voltage engineers use the line-charge result every day. The field at the surface of a wire of radius $a$ is $\frac{\lambda}{2\pi\varepsilon_0a}$, so thin wires reach air's breakdown field (about 3 MV/m) first: that's the hiss and glow of corona discharge on power lines, and why the highest-voltage lines bundle several conductors per phase so they act like one fat wire. Coaxial cables are rated with the same formula at their inner conductor. And because gravity is also an inverse-square force, the same law, with $G$ in place of $k$, shows that a planet pulls like a point mass on anything outside it.