Conductors and Capacitance

Equilibrium Forces the Field Out

A conductor in electrostatic equilibrium has no moving charge, and that single fact settles everything else. If any field remained inside, free charges would move, so $\mathbf{E} = \mathbf{0}$ throughout the interior. Consequently:

• All excess charge sits on the surface, since $\nabla \cdot \mathbf{E} = 0$ inside forces $\rho = 0$ there.
• The whole conductor is an equipotential: with no interior field there is no potential difference between any two of its points.
• The surface field is perpendicular to the surface, and equals $\sigma/\varepsilon_0$ just outside. A tangential component would drive surface currents.

The charge experiences an outward electrostatic pressure $\sigma^2/2\varepsilon_0$, the reason a heavily charged conductor can physically tear.

Capacitance measures how much charge a conductor system holds per volt: $C = Q/V$, in farads. It depends only on geometry and the medium, never on how much charge you happen to have put there. For a parallel-plate capacitor

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$$

Combinations follow from which quantity the elements share:

Note this is the opposite of resistors, and for the same reason: series capacitors share charge while series resistors share current.

With several conductors, electrostatic influence means each one's potential depends on all the charges, through a matrix of capacitance and influence coefficients. A conductor fully enclosing another shields it completely, the basis of the Faraday cage.

Common pitfall: thinking capacitance grows when you add charge. $C$ is a fixed property of the geometry: doubling $Q$ doubles $V$ and leaves $C$ untouched. It changes only if you alter the plate area, the separation, or the dielectric.