Gauss's law becomes a tool for finding fields when the charge is spherically symmetric: a point charge, a charged shell, a ball, or layers of these. Symmetry says the field points straight out (or in) and has the same size everywhere on a sphere centered on the charge. On such a sphere the flux integral is just $E$ times the area.
Why it works
Draw a Gaussian sphere of radius $r$ through the point you care about. The field is radial and the same size all over it, so
$$\oint \vec{E}\cdot d\vec{A} = E\,(4\pi r^2) = \frac{q_{\text{enc}}}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{k\,q_{\text{enc}}}{r^2}.$$
Every problem in this topic comes down to one question: how much charge is inside radius $r$? Charge farther out than $r$ contributes nothing at radius $r$.
Key idea
Shell of charge $Q$, radius $R$: $E = 0$ inside, $E = kQ/r^2$ outside.
Uniform ball: inside, only the fraction $r^3/R^3$ is enclosed, so $E = \dfrac{kQr}{R^3}$, growing in step with $r$; outside, $E = kQ/r^2$.
Uniform ball: inside, only the fraction $r^3/R^3$ is enclosed, so $E = \dfrac{kQr}{R^3}$, growing in step with $r$; outside, $E = kQ/r^2$.
Worked examples
Worked example: Inside and outside a ball
A ball of radius $6.0\ \text{cm}$ holds $12\ \text{nC}$ spread evenly. Find the field $3.0\ \text{cm}$ and $12\ \text{cm}$ from the center.
- At $3.0\ \text{cm}$ (inside), the enclosed fraction is $(3/6)^3 = 1/8$. $$E = \frac{kQr}{R^3} = \frac{(8.99 \times 10^{9})(12 \times 10^{-9})(0.030)}{(0.060)^3} = 1.50 \times 10^{4}\ \text{N/C}$$
- At $12\ \text{cm}$ (outside), all of $Q$ is enclosed: $$E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^{9})(12 \times 10^{-9})}{(0.12)^2} = 7490\ \text{N/C}$$
- At the surface both formulas give the same value, $kQ/R^2 = 3.00 \times 10^{4}\ \text{N/C}$, so the field is continuous there.
Answer: $1.50 \times 10^{4}\ \text{N/C}$ and $7490\ \text{N/C}$
Worked example: A charge inside a charged shell
A $+3.0\ \text{nC}$ point charge sits at the center of a thin shell of radius $5.0\ \text{cm}$ carrying $-5.0\ \text{nC}$. Find the field at $3.0\ \text{cm}$ and at $10\ \text{cm}$.
- At $3.0\ \text{cm}$ the Gaussian sphere encloses only the point charge: $E = \dfrac{(8.99 \times 10^{9})(3.0 \times 10^{-9})}{(0.030)^2} = 3.00 \times 10^{4}\ \text{N/C}$, outward.
- At $10\ \text{cm}$ it encloses $3.0 - 5.0 = -2.0\ \text{nC}$: $E = \dfrac{(8.99 \times 10^{9})(2.0 \times 10^{-9})}{(0.10)^2} = 1800\ \text{N/C}$, pointing inward because the enclosed charge is negative.
Answer: $3.00 \times 10^{4}\ \text{N/C}$ outward; $1800\ \text{N/C}$ inward
Worked example: A density that grows outward
A ball of radius $R = 10\ \text{cm}$ has $\rho = \rho_0 r/R$ with $\rho_0 = 20\ \mu\text{C/m}^3$. Find $E$ at $r = 5.0\ \text{cm}$.
- Enclosed charge, by thin shells: $$q(r) = \int_0^r \rho_0\frac{s}{R}\,4\pi s^2\,ds = \frac{\pi\rho_0 r^4}{R}$$
- Gauss: $E(4\pi r^2) = \dfrac{\pi\rho_0 r^4}{\varepsilon_0 R}$, so $E = \dfrac{\rho_0 r^2}{4\varepsilon_0 R}$.
- $$E = \frac{(20 \times 10^{-6})(0.050)^2}{4(8.85 \times 10^{-12})(0.10)} = 1.41 \times 10^{4}\ \text{N/C}$$
Answer: $E \approx 1.41 \times 10^{4}\ \text{N/C}$
Common mistake
"The field inside a charged ball is zero, like inside a shell." A shell has no charge inside it, so a Gaussian sphere there encloses nothing. A ball has charge all the way in, so the enclosed charge, and the field, grow as you move out: $E = kQr/R^3$. Only at the very center is $E = 0$. Check with the first example: halfway out, the field is half the surface value, not zero.