Conductors in electrostatic equilibrium

Last week you applied Gauss’s law to insulators; the field was nonzero even inside the charge. Conductors behave differently, because they contain free electrons, and these electrons keep moving as long as they feel a force. Most exam questions on this topic are concentric sphere-and-shell problems: “how much charge is on the inner surface, how much on the outer surface, what is the field at point $r$?” If you know the four properties and put the Gaussian surface in the right place, you won’t miss this question.

Golden rules:

1. Inside a conductor $E = 0$. For every conductor in equilibrium, whatever its shape.
2. Excess charge goes to the surface. There is no net charge inside a conductor.
3. Just outside the surface the field is perpendicular to the surface and $E = \sigma/\varepsilon_0$.
4. Charge is denser at a pointed place. $\sigma$ is larger where the radius of curvature is smaller.

Electrostatic equilibrium

When you put charge on a conductor or place it in an external field, the free electrons move for a short time (much less than a nanosecond). Then they stop. There is no net motion of charge inside any more: this is called electrostatic equilibrium. During that short time the electrons take on a distribution such that all four properties hold at once.

1. The field inside is zero

Suppose $E \neq 0$ at some point inside. The free electrons there feel the force $\vec F = -e\vec E$ and move. But in equilibrium there is no motion. Contradiction. So $E = 0$.

Think of a metal slab placed in an external field. If the field points right, the electrons shift left: the left face becomes negative and the right face becomes positive. These induced charges produce a field pointing left inside. The electrons keep shifting until this field exactly cancels the external field.

2. Excess charge is on the surface

Draw a Gaussian surface inside the conductor, just below the surface. At every point $E = 0$, so the flux is zero, so $q_{in} = 0$. You can bring this Gaussian surface as close to the surface as you like. So all the excess charge of the conductor is on the surface.

3. Just outside the surface the field is perpendicular and equals $\sigma/\varepsilon_0$

If the field had a component parallel to the surface, the electrons on the surface would slide along it. In equilibrium they don’t, so the field is perpendicular to the surface.

For the magnitude, choose a small box (pillbox) through the surface. On the cap inside, $E = 0$ (inside the metal). On the side faces the field is parallel. The flux comes only from the outer cap:

$EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{\varepsilon_0}$

Don’t confuse it with $\sigma/(2\varepsilon_0)$ for the insulating sheet. For the insulating sheet the field went out in both directions, and there was flux through both caps of the box. In a conductor the inside is zero, and all the flux comes out through one cap.

PITFALL · Two “sheet” formulas. Charged insulating sheet: $\sigma/(2\varepsilon_0)$. Just outside a conductor’s surface: $\sigma/\varepsilon_0$. Tell which one is asked from the words “nonconducting” and “conducting”.

4. Charge is denser at a pointed place

In a spherical conductor, $\sigma$ is the same everywhere because of symmetry. But in a shape like an egg or a pear, the charge collects at the pointed end where the radius of curvature is small. There $\sigma$ is large, and so is the field outside. The full explanation comes in the potential topic (the surface of a conductor is an equipotential). For now just know the result: lightning rods are pointed, and corona discharge occurs at sharp points on high-voltage lines.

Example 1. A conducting sphere of radius 0.10 m carries $+4.0\ \mu$C.

σ = Q / (4πR²) = 4.0e-6 / (4π·0.010) = 3.18e-5 C/m²
E_surface = σ / ε0 = 3.18e-5 / 8.85e-12 = 3.60e6 N/C
Check: kQ/R² = (8.99e9)(4.0e-6)/(0.10)² = 3.60e6 N/C  ✓

Outside a conducting sphere, the field is the same as that of a point charge with the same charge. The difference is inside: 0 in a conductor, $kQr/R^3$ in a uniform insulator.

Conductors with a cavity

Let there be a cavity inside the conductor, with a charge $q$ in the cavity. Draw a Gaussian surface that passes through the metal and encloses the cavity. In the metal $E = 0$, so the net charge inside this surface must be zero. To cancel the $q$ in the cavity, $-q$ is induced on the wall of the cavity (the inner surface).

If the conductor’s own total charge is $Q$, then by conservation the outer surface has:

$q_{\text{outer}} = Q - q_{\text{inner}} = Q + q$

If there is no charge in the cavity, the inner surface is uncharged and the field in the cavity is zero. This is electrostatic shielding: the inside of a closed metal box is protected from external fields. This is why people inside a car are not harmed when lightning strikes it, why a phone has no signal in an elevator, and why sensitive devices are put in metal boxes. A cage instead of a solid box is also enough (Faraday cage).

Concentric sphere and shell problems

The classic exam question. The method is always the same:

1. List the surfaces: the surface of the inner sphere ($a$), the inner surface of the shell ($b$), the outer surface of the shell ($c$).
2. Inner surface of the shell: put a Gaussian surface through the metal and use $q_{in} = 0$. The result is the negative of the inner sphere’s charge.
3. Outer surface of the shell: the total charge of the shell minus the inner surface.
4. Field in each region: $E = kq_{in}/r^2$; if you are inside the metal, 0.

Example 2. A conducting sphere of radius $a = 3.0$ cm carries $+2.0\ \mu$C. Around it is a conducting shell with inner radius $b = 6.0$ cm and outer radius $c = 9.0$ cm; the net charge of the shell is $-5.0\ \mu$C.

Surface charges:
  r = a (surface of the sphere):       +2.0 μC
  r = b (inner surface of the shell):  −2.0 μC   (so that q_in = 0 in the metal)
  r = c (outer surface of the shell):  −5.0 − (−2.0) = −3.0 μC

Regions:
  r = 2 cm  (metal of the inner sphere):  E = 0
  r = 5 cm  (gap, q_in = +2.0 μC):
            E = (8.99e9)(2.0e-6)/(0.05)² = 7.19e6 N/C, outward
  r = 7 cm  (metal of the shell):         E = 0
  r = 12 cm (outside, q_in = +2.0 − 5.0 = −3.0 μC):
            E = (8.99e9)(3.0e-6)/(0.12)² = 1.87e6 N/C, inward

Density of the outer surface: σ_c = −3.0e-6 / (4π·0.09²) = −2.95e-5 C/m²

TIP · Check. In every region that falls inside the metal, the field must come out zero. If it doesn’t, you wrote the charge of some surface wrongly.

Thin conducting plate

If you give a charge $Q$ to a large, thin, isolated metal plate, the charge is shared equally between the two faces (each face has area $A$):

$\sigma_{\text{face}} = \frac{Q}{2A}, \qquad E_{\text{outside}} = \frac{\sigma_{\text{face}}}{\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$

Example 3. $A = 0.25\ \text{m}^2$, $Q = 3.0\ \mu$C: $\sigma_{\text{face}} = 6.0 \times 10^{-6}$ C/m², $E = 6.78 \times 10^{5}$ N/C. If the same charge were spread uniformly over an insulating plate ($\sigma = Q/A$, $E = \sigma/(2\varepsilon_0)$), it would give numerically the same result. The field outside is the same; the difference is inside the conductor and in where the charge sits.

Experimental check of Gauss’s law

Gauss’s law and Coulomb’s inverse-square law are two sides of the same thing. If the inverse-square law were not exactly $1/r^2$, the field inside a charged conductor would not be exactly zero, and some of the charge would stay on the inner surface. That is why the inverse-square law is tested most precisely by measuring whether any charge is left inside the conductor.

The classic experiment is Faraday’s ice-pail experiment:

1. A small charged metal ball is lowered, without touching, into an insulated neutral metal can. $-q$ is induced on the inner surface of the can and $+q$ on the outer surface. An electroscope connected to the outer surface deflects.
2. The ball is touched to the inner wall of the can. The electroscope deflection does not change.
3. The ball is taken out and examined: it has no charge left.

All of the ball’s charge has passed to the can, and all of it went to the outer surface. In modern versions, the charge left on the inner sphere has been measured so precisely that the deviation of the exponent in $1/r^n$ from 2 is found to be smaller than about $10^{-16}$.

Common mistakes