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# 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

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