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 \(r_b\).

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 \(4\pi {r_b}^2\).