2.2B: Spherical Charge Distributions

Outside any spherically-symmetric charge distribution, the field is the same as if all the charge were concentrated at a point in the centre, and so, then, is the potential. Thus

$$V=\frac{Q}{4\pi\epsilon_0 r}.\tag{2.2.3}$$

Inside a hollow spherical shell of radius a and carrying a charge $Q$ the field is zero, and therefore the potential is uniform throughout the interior, and equal to the potential on the surface, which is

$$V=\frac{Q}{4\pi\epsilon_0 a}.\tag{2.2.4}$$

A solid sphere of radius a bearing a charge $Q$ that is uniformly distributed throughout the sphere is easier to imagine than to achieve in practice, but, for all we know, a proton might be like this (it might be – but it isn’t!), so let’s calculate the field at a point P inside the sphere at a distance $(r < a)$ from the centre. See Figure $II.1$

We can do this in two parts. First the potential from the part of the sphere “below” P. If the charge is uniformly distributed throughout the sphere, this is just $\frac{Q_r}{4\pi\epsilon_0 r}$. Here $Q_r$ is the charge contained within radius $r$, which, if the charge is uniformly distributed throughout the sphere, is $Q(r^3/a^3)$. Thus, that part of the potential is $\frac{Qr^2}{4\pi\epsilon_0 a^3}$.

$\text{FIGURE II.1}$

Next, we calculate the contribution to the potential from the charge “above” P. Consider an elemental shell of radii $x ,\, x + δx$. The charge held by it is $\delta Q = \frac{4\pi x^2 \delta x}{\frac{4}{3}\pi a^3}\times Q=\frac{3Qx^2 \delta x}{a^3}$. The contribution to the potential at P from the charge in this elemental shell is $\frac{\delta Q}{4\pi\epsilon_0 x}=\frac{3Qx\delta x}{4\pi\epsilon_0 a^3}$. The contribution to the potential from all the charge “above” P is $\frac{3Q}{4\pi\epsilon_0 a^3}\int_r^a x\,dx=\frac{3Q(a^2-r^2)}{4\pi\epsilon_0 2a^3}$. Adding together the two parts of the potential, we obtain

$$V=\frac{Q}{8\pi\epsilon_0 a^3}(3a^2-r^2).\tag{2.2.5}$$