$A$ spherical capacitor consists of two concentric spherical conductors,held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by $C = \frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1} - r_{2}}$,where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres,respectively.

(N/A) Let the radius of the outer shell be $r_{1}$ and the radius of the inner shell be $r_{2}$.
The inner shell is given a charge $+Q$ and the outer shell is given a charge $-Q$.
The potential at the surface of the inner shell due to its own charge is $V_{inner} = \frac{Q}{4 \pi \epsilon_{0} r_{2}}$.
The potential at the surface of the inner shell due to the charge on the outer shell is $V_{outer} = -\frac{Q}{4 \pi \epsilon_{0} r_{1}}$.
The potential difference $V$ between the two shells is $V = V_{inner} - V_{outer} = \frac{Q}{4 \pi \epsilon_{0} r_{2}} - \frac{Q}{4 \pi \epsilon_{0} r_{1}}$.
$V = \frac{Q}{4 \pi \epsilon_{0}} \left( \frac{1}{r_{2}} - \frac{1}{r_{1}} \right) = \frac{Q(r_{1} - r_{2})}{4 \pi \epsilon_{0} r_{1} r_{2}}$.
The capacitance $C$ is defined as $C = \frac{Q}{V}$.
Substituting the value of $V$,we get $C = \frac{Q}{\frac{Q(r_{1} - r_{2})}{4 \pi \epsilon_{0} r_{1} r_{2}}} = \frac{4 \pi \epsilon_{0} r_{1} r_{2}}{r_{1} - r_{2}}$.
Hence,the capacitance is proved.