Field Outside Uniformly Charged Spherical Shell

Field at r > R due to a shell of radius R carrying total charge Q. Behaves as if all charge is concentrated at the centre.

Setup

A thin spherical shell of radius $R$ carries total charge $Q$ uniformly distributed on its surface (surface charge density $\sigma = Q/4\pi R^2$).

For $r > R$, spherical symmetry tells us $\vec{E}$ is radial and has the same magnitude at all points on a concentric sphere of radius $r$.

Gaussian surface

Choose a concentric spherical surface of radius $r > R$.

$$\oint_S \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2$$

Enclosed charge $= Q$ (entire shell is inside).

By Gauss's law:

$$E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}$$ $$\boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2} \quad (r > R)}$$

Shell theorem

The field outside a uniformly charged shell is exactly the same as if all the charge were concentrated at the centre. This is the electrostatic shell theorem — the three-dimensional analogue of Newton's shell theorem for gravity.