Consider two concentric spherical shells separated by vacuum as shown in Figure 1. The inner shell has charge \(+Q\) and the outer shell has charge \(-Q\). It means the charge on the capacitor is \(Q\) (note that it is a common practice to represent the magnitude of charge on either conductor as the charge on the capacitor, otherwise you know the charge of a capacitor is zero). The outer radius of the inner shell is \(r_a\) and the inner radius of the outer shell is \(r_b\).

Figure 1 A spherical capacitor; the electric field between the conductors is due to the inner conducting spherical shell.

The electric field due to the outer shell has no effect on electric field between the shells. We know from the application of Gauss's law that the electric field inside a conducting sphere is zero. It means the electric field between the shells is due to inner shell with charge \(+Q\). So take any point between the shells at distance \(r\) from the centre and the electric field at that point due to the inner shell is

\[E = k\frac{Q}{r^2}\]

The electric field is radially outward and the small potential difference for a very small displacement \(d\vec r\) along the radial line is \(\vec E \cdot d\vec r\). Now the potential difference between the plates is determined by integrating this small potential difference from \(r_a\) to \(r_b\). Note that both \(\vec E\) and \(d\vec r\) are parallel, so

\[\begin{align*} {V_{ab}} &= \int\limits_{{r_a}}^{{r_b}} {\vec E \cdot d\vec {r} } \\ {\rm{or, }}\quad {V_{ab}} &= \int\limits_{{r_a}}^{{r_b}} {Edr = kQ\frac{{{r_b} - {r_a}}}{{{r_a}{r_b}}}} \end{align*}\]

Now the capacitance \(C\) is \(C = Q/V_{ab}\) and

\[C = \frac{1}{k}\frac{{{r_a}{r_b}}}{{{r_b} - {r_a}}} = {\epsilon_0}\frac{{4\pi {r_a}{r_b}}}{{{r_b} - {r_a}}}\]

In the above expression of the capacitance the quantity \(4\pi r_a r_b\) is the geometric mean of areas \(4\pi {r_a}^2\) and \(4\pi {r_b}^2\).