Field of a Spherical Charge

Chapter 1 — Electrostatics

Any spherically symmetric charge distribution — a thin shell, a uniform ball, a nested sequence of shells — produces a field outside that looks identical to a point charge at the center. Inside, the field depends only on the charge interior to your radius. Both facts pop out of Gauss's law in a couple of lines.

Thin spherical shell

A shell of radius \(R\) carrying total charge \(Q\). Spherical symmetry: \(\vec E = E(r)\hat r\). Pick a concentric sphere of radius \(r\).

Outside (\(r > R\)): \(Q_{\text{enc}} = Q\). Gauss: \(E\cdot 4\pi r^2 = Q/\epsilon_0\), so

\[\vec E_{\text{out}} = \frac{kQ}{r^2}\hat r.\]

Inside (\(r < R\)): \(Q_{\text{enc}} = 0\). Gauss: \(E\cdot 4\pi r^2 = 0\), so

\[\vec E_{\text{in}} = 0.\]

The field jumps discontinuously from \(0\) to \(kQ/R^2\) across the shell. This is the origin of the surface-charge discontinuity you'll see again at conductor surfaces.

Uniformly charged solid ball

Total charge \(Q\) spread uniformly through a ball of radius \(R\). Charge density \(\rho = Q/(\tfrac43\pi R^3)\).

Outside (\(r > R\)): same as a point charge: \(E = kQ/r^2\).

Inside (\(r < R\)): \(Q_{\text{enc}} = \rho\cdot \tfrac43\pi r^3 = Q(r/R)^3\). Then

\[E(r) = \frac{kQ_{\text{enc}}}{r^2} = \frac{kQ r}{R^3}.\]

Field grows linearly with \(r\) from \(0\) at the center to \(kQ/R^2\) at the surface, then falls as \(1/r^2\) outside.

General spherically symmetric distribution

For any \(\rho(r)\), the field at radius \(r\) is

\[E(r) = \frac{k}{r^2}\int_0^r 4\pi r'^2 \rho(r')\,dr' = \frac{k Q_{\text{enc}}(r)}{r^2}.\]

Two punchlines: (1) only the charge interior to \(r\) contributes to \(\vec E\) at \(r\); (2) exterior shells contribute nothing at interior points — a hollow spherical shell of charge shields its interior from its own field. (Nonspherical shells do not have this property — a hollow cube of charge has nonzero field inside.)

Practice Problems