A metal is full of electrons free to move. Put extra charge on it, or hold it in a field, and those electrons rearrange within a tiny fraction of a second, then stop. That final, still state is electrostatic equilibrium. With Gauss's law and the pillbox, three facts about it follow quickly.
Key idea
In a conductor in electrostatic equilibrium:
1. $E = 0$ everywhere inside the metal.
2. Any excess charge sits on the surface.
3. Just outside, $\vec{E}$ is perpendicular to the surface, with $E = \sigma/\varepsilon_0$.
1. $E = 0$ everywhere inside the metal.
2. Any excess charge sits on the surface.
3. Just outside, $\vec{E}$ is perpendicular to the surface, with $E = \sigma/\varepsilon_0$.
Why it works
1: if there were a field inside, the free electrons would feel a force and move. "Nothing moves" therefore means $E = 0$.
2: draw a Gaussian surface anywhere inside the metal, just under the surface. $E = 0$ on it, so its flux is zero and it encloses zero net charge. The excess has nowhere to be but the surface.
3: a field along the surface would push charges along it, so the outside field must be perpendicular. A tiny pillbox with one cap outside and one inside the metal gets flux only through the outer cap: $EA = \sigma A/\varepsilon_0$. That is twice a lone sheet's $\sigma/(2\varepsilon_0)$, because here all the field goes out one side.
2: draw a Gaussian surface anywhere inside the metal, just under the surface. $E = 0$ on it, so its flux is zero and it encloses zero net charge. The excess has nowhere to be but the surface.
3: a field along the surface would push charges along it, so the outside field must be perpendicular. A tiny pillbox with one cap outside and one inside the metal gets flux only through the outer cap: $EA = \sigma A/\varepsilon_0$. That is twice a lone sheet's $\sigma/(2\varepsilon_0)$, because here all the field goes out one side.
Worked examples
Worked example: Field from surface charge
A flat part of a charged metal plate has $\sigma = 20\ \text{nC/m}^2$. Find the field just outside.
- $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{20 \times 10^{-9}}{8.85 \times 10^{-12}} = 2260\ \text{N/C}$, perpendicular to the plate.
Answer: $E \approx 2260\ \text{N/C}$
Worked example: A charge in a cavity
A $+3.0\ \text{nC}$ point charge sits at the center of a hollow metal shell that carries net charge $-5.0\ \text{nC}$. How is the shell's charge divided between its inner and outer surfaces?
- Draw a Gaussian sphere inside the metal. $E = 0$ there, so it encloses zero net charge: the inner surface must carry $-3.0\ \text{nC}$ to cancel the point charge.
- The shell's total is $-5.0\ \text{nC}$, so the outer surface carries $-5.0 - (-3.0) = -2.0\ \text{nC}$.
Answer: Inner $-3.0\ \text{nC}$, outer $-2.0\ \text{nC}$
Worked example: Fields in each region
Same setup, with the shell's inner radius $5.0\ \text{cm}$ and outer radius $8.0\ \text{cm}$. Find the field at $3.0$, $6.0$ and $12\ \text{cm}$.
- At $3.0\ \text{cm}$ (in the cavity): only the point charge is enclosed, $E = kq/r^2 = 3.00 \times 10^{4}\ \text{N/C}$, outward.
- At $6.0\ \text{cm}$ (in the metal): $E = 0$.
- At $12\ \text{cm}$ (outside): $q_{\text{enc}} = 3.0 - 5.0 = -2.0\ \text{nC}$, so $E = 1250\ \text{N/C}$, pointing inward.
Answer: $3.00 \times 10^{4}\ \text{N/C}$, $0$, $1250\ \text{N/C}$ inward
Common mistake
"Extra charge spreads through the whole inside of a solid metal ball." In an insulator charge can stay where you put it, but in a metal it cannot: like charges push each other until every bit of excess sits on the outer surface. Check with Gauss's law: any surface drawn inside the metal has $E = 0$ on it, so it encloses no net charge.
Fact 1 also means a closed metal box shields its inside from static fields outside: charges on the box's surface rearrange until the field inside the metal, and in the empty space within, is zero. This is the idea behind a Faraday cage.