Obtain an expression for the electric intensity at a point outside a uniformly charged spherical shell.

Answer: \(E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2} = \dfrac{\sigma R^2}{\varepsilon_0 r^2}\)

Consider a thin spherical shell of radius \(R\) with charge \(q\) uniformly distributed (surface charge density \(\sigma = \dfrac{q}{4\pi R^2}\)). Let P be a point outside at distance \(r\) \((r \gt R)\) from the centre O.

[Figure: charged shell with centre O and a concentric dotted Gaussian sphere of radius \(r\) passing through P; \(\vec{E}\) and \(d\vec{s}\) both radially outward at P.]

Draw a concentric spherical Gaussian surface of radius \(r\) through P. By symmetry \(\vec{E}\) is radial and has the same magnitude at every point of this surface, and it is parallel to \(d\vec{s}\) (\(\theta = 0\)). The flux through the Gaussian surface is

By Gauss’s theorem \(\phi = \dfrac{q}{\varepsilon_0}\). Hence

Using \(q = 4\pi R^2\sigma\), \(E = \dfrac{\sigma R^2}{\varepsilon_0 r^2}\). So for an external point the shell behaves as if its whole charge were concentrated at its centre.