Gauss's Law

Wrap any closed surface around some charges and add up the electric flux leaving it (topic 8.5 defined flux). The total depends on exactly one thing: the net charge inside. That fact is Gauss's law. It is true for every closed surface and every arrangement of charge, and for charge spread with enough symmetry it replaces the field integrals of topic 8.4 with one line of algebra.

What the law says

A Gaussian surface is a closed surface: it encloses a volume with no gaps, like a balloon or a sealed box, rather than an open sheet like a flag. It is imaginary, so you can draw it anywhere, including straight through charged material. As in topic 8.5, the area vector $d\vec{A}$ of each small patch points outward, away from the enclosed volume.

Two details decide most questions about the law.

• The field in the integral is the total field at each point of the surface, produced by every charge, inside and outside. Charges outside the surface change $\vec{E}$ on the surface.
• Charges outside contribute zero net flux. Every field line from an outside charge that enters the surface leaves it somewhere else, so its inward flux and outward flux cancel. That is why only $q_{\text{enc}}$ appears on the right.

For example, put $+4.0\ \text{nC}$ and $-1.5\ \text{nC}$ inside a lumpy closed surface and $+6.0\ \text{nC}$ just outside it. The net flux is

no matter where inside the charges sit or what shape the surface has. The $+6.0\ \text{nC}$ charge reshapes the field on the surface but adds nothing to the total. A negative $q_{\text{enc}}$ gives negative flux: more field enters the surface than leaves.

Why the size and shape of the surface don't matter

Start with a point charge $q$ at the center of a sphere of radius $r$. On the sphere the field has magnitude $\frac{q}{4\pi\varepsilon_0 r^2}$ and points straight outward, parallel to $d\vec{A}$ at every patch, so $\vec{E}\cdot d\vec{A} = E\,dA$ and $E$ is the same everywhere on the sphere:

The radius cancels. A bigger sphere has a weaker field on it, falling as $\frac{1}{r^2}$, but more area, growing as $r^2$, and the two changes offset exactly. Gauss's law is the inverse-square law in disguise: if the field fell off as $\frac{1}{r^3}$, the flux through a sphere would shrink as $\frac{1}{r}$ and no such law could hold.

For a surface of any other shape, think in field lines. The number of lines leaving a charge is proportional to the charge, and every line that starts inside a closed surface must cross it outward (on net) to get out, whatever the surface looks like. So the flux through any closed surface around $q$ is the same as through the sphere, $\frac{q}{\varepsilon_0}$. With several charges, the fields add, so the fluxes add, and each outside charge contributes zero.

The practical consequence: if you grow or shrink a Gaussian surface while the charge it encloses stays the same, the total flux does not change, even though the field on the surface does. If the surface cuts through a charged object, growing it encloses more charge and the flux grows with it.

Gauss's law is also the first of Maxwell's equations, the four equations that together describe all of classical electricity and magnetism: Gauss's law for electric fields, Gauss's law for magnetic fields (topic 12.1), Faraday's law (topic 13.2), and Ampère's law with Maxwell's addition (topic 12.4). Coulomb's law only describes charges at rest; Gauss's law stays true even when charges move.

Picking a surface that makes the integral easy

Gauss's law always holds, but it gives you $E$ only when you can pull $E$ out of the integral. That requires a surface built from pieces of two kinds:

• pieces where $\vec{E}$ is perpendicular to the surface (parallel to $d\vec{A}$) and has the same magnitude everywhere, so that piece's flux is just $EA$;
• pieces where $\vec{E}$ lies along the surface, so $\vec{E}\cdot d\vec{A} = 0$ and the piece carries no flux.

Symmetry tells you the field's direction before you calculate anything. For a ball whose charge density depends only on distance from the center, any rotation about the center leaves the charge looking the same, so the field cannot favor any direction other than radial, and its magnitude can depend only on $r$. For a very long cylinder with charge depending only on distance from the axis, the field points straight away from the axis (a component along the axis would have to pick one end over the other). For a large flat sheet, the field is perpendicular to the sheet and has the same magnitude at equal distances on either side.

The sheet result comes from setting the flux through the two end faces, $2EA$, equal to $\frac{\sigma A}{\varepsilon_0}$, where $\sigma A$ is the charge on the patch of sheet inside the box. The distance from the sheet never appears, so the field of an infinite sheet is the same at every distance: its field lines are parallel and do not spread out. For a line charge $\lambda$, the cylinder gives $E = \frac{\lambda}{2\pi\varepsilon_0 r}$, the same result topic 8.4 reaches by integrating along the whole line.

Inside a uniformly charged solid ball of total charge $Q$ and radius $R$, a Gaussian sphere of radius $r < R$ encloses only the fraction $\frac{r^3}{R^3}$ of the volume, so $q_{\text{enc}} = Q\frac{r^3}{R^3}$ and

The field grows linearly from zero at the center to its largest value at the surface; outside, it falls as $\frac{1}{r^2}$, exactly as if all the charge sat at the center.

The exam asks you to use Gauss's law to calculate fields only for point charges and for spherical, cylindrical and planar symmetry. For a finite rod, a cube or a disk, the law is still true, but no closed surface has a field of constant magnitude perpendicular to it, so it cannot give you $E$; use topic 8.4's integration instead. Topic 10.1 applies Gauss's law to conductors.

Finding the enclosed charge from a density

Charge spread over a region is described by a density: $\lambda$ (coulombs per meter) along a line, $\sigma$ (coulombs per square meter) over a surface, or $\rho$ (coulombs per cubic meter) through a volume. If the density is uniform, multiply it by the length, area or volume. If it varies, add up thin pieces on which it is nearly constant:

The trick is to choose pieces whose every point has the same density. When $\rho$ depends only on $r$:

• for a sphere, use thin spherical shells, $dV = 4\pi r^2\,dr$;
• for a cylinder of length $\ell$, use thin cylindrical shells, $dV = 2\pi r\ell\,dr$;
• for a flat disk with $\sigma$ depending on $r$, use thin rings, $dA = 2\pi r\,dr$.

Two short examples. A rod along the $x$-axis from $0$ to $L$ with $\lambda = \beta x^2$ carries $Q = \int_0^L \beta x^2\,dx = \frac{\beta L^3}{3}$. A disk of radius $R$ with $\sigma = \sigma_0\frac{r}{R}$ carries $Q = \int_0^R \sigma_0\frac{r}{R}\,2\pi r\,dr = \frac{2\pi\sigma_0 R^2}{3}$.

For Gauss's law you usually need the charge inside the Gaussian surface, not the total. Then the upper limit of the integral is the radius of the Gaussian surface (if it lies inside the charge) or the edge of the charge (if the surface lies outside it).

Worked examples

Worked exampleExample 3: a nonuniform cylinder (free-response style)

A very long solid insulating cylinder of radius $R$ has a volume charge density that falls off from its axis as

$$\rho(r) = \rho_0\left(1 - \frac{r}{R}\right) \quad \text{for } r \le R,$$

where $\rho_0$ is a positive constant and $r$ is the distance from the axis. Express answers in terms of $\rho_0$, $R$, $r$ and physical constants.

(a) Derive an expression for the charge per unit length $\lambda$ of the cylinder.

(b) Using Gauss's law, derive an expression for the magnitude of the electric field at a distance $r < R$ from the axis.

(c) Derive an expression for the magnitude of the field at a distance $r > R$.

(d) Sketch $E$ as a function of $r$ from $r = 0$ to $r = 3R$. A student claims the field is strongest at the surface of the cylinder. Is the student correct? Justify your answer.

Show solution

(a) The density depends only on $r$, so slice a length $\ell$ of the cylinder into thin cylindrical shells of radius $r'$, thickness $dr'$ and volume $dV = 2\pi r'\ell\,dr'$, each with uniform density:

$$q = \int_0^R \rho_0\left(1 - \frac{r'}{R}\right)2\pi r'\ell\,dr' = 2\pi\rho_0\ell\left(\frac{R^2}{2} - \frac{R^2}{3}\right) = \frac{\pi\rho_0 R^2\ell}{3}.$$

Dividing by $\ell$, $\lambda = \dfrac{\pi\rho_0 R^2}{3}$.

(b) By cylindrical symmetry, $\vec{E}$ points radially away from the axis and depends only on $r$. Take a coaxial Gaussian cylinder of radius $r < R$ and length $\ell$. The field is perpendicular to its curved side with constant magnitude and lies along its end caps, so the flux is $E(2\pi r\ell)$. The enclosed charge is the same integral with upper limit $r$:

$$q_{\text{enc}} = \int_0^r \rho_0\left(1 - \frac{r'}{R}\right)2\pi r'\ell\,dr' = 2\pi\rho_0\ell\left(\frac{r^2}{2} - \frac{r^3}{3R}\right).$$

Gauss's law, $E(2\pi r\ell) = \frac{q_{\text{enc}}}{\varepsilon_0}$, gives

$$E = \frac{\rho_0}{\varepsilon_0}\left(\frac{r}{2} - \frac{r^2}{3R}\right).$$

(c) For $r > R$ the same Gaussian cylinder encloses all the charge in its length, $\lambda\ell$:

$$E(2\pi r\ell) = \frac{\pi\rho_0R^2\ell}{3\varepsilon_0} \quad\Rightarrow\quad E = \frac{\rho_0R^2}{6\varepsilon_0 r}.$$

As a check, both expressions give $E = \frac{\rho_0R}{6\varepsilon_0}$ at $r = R$. With no surface charge, the field must be continuous there.

(d) The sketch starts at $E = 0$ on the axis (no enclosed charge), rises, reaches a maximum inside the cylinder, comes down to $\frac{\rho_0R}{6\varepsilon_0}$ at $r = R$ with no jump, and then decreases as $\frac{1}{r}$, so that at $r = 2R$ it is half its value at $R$ and at $3R$ a third.

The student is not correct. Inside, $\frac{dE}{dr} = \frac{\rho_0}{\varepsilon_0}\left(\frac{1}{2} - \frac{2r}{3R}\right)$, which is zero at $r = \frac{3}{4}R$, positive before it and negative after it. So the field is largest at $r = \frac{3}{4}R$, where $E = \frac{3\rho_0R}{16\varepsilon_0}$, which is greater than the surface value $\frac{\rho_0R}{6\varepsilon_0}$. Physically, past $r = \frac{3}{4}R$ the thin outer layers add so little charge that the enclosed charge grows more slowly than the area $2\pi r\ell$ it is spread over.

Where students lose points

On the exam

In a Gauss's law derivation, show each step the reasoning needs: state the law, name the Gaussian surface and say why $\vec{E}$ is constant and perpendicular on it (and zero flux through the other parts), compute $q_{\text{enc}}$ (with an integral when the density varies), and solve for $E$. Sketches of $E$ against $r$ are graded on behavior: zero or linear growth inside, the right falloff outside ($\frac{1}{r^2}$ for spheres, $\frac{1}{r}$ for cylinders, constant for planes), and continuity at a surface with no surface charge. Factor-of-change questions ("the radius doubles; what happens to $E$?") come straight from those dependences. A calculator is allowed throughout the exam.