**Question:**

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36).

Show that the capacitance of a spherical capacitor is given by

$C=\frac{4 \pi \in_{0} r_{1} r_{2}}{r_{1}-r_{2}}$

where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres, respectively.

**Solution:**

Radius of the outer shell = *r*1

Radius of the inner shell = *r*2

The inner surface of the outer shell has charge +*Q*.

The outer surface of the inner shell has induced charge −*Q*.

Potential difference between the two shells is given by,

$V=\frac{Q}{4 \pi \epsilon_{0} r_{2}}-\frac{Q}{4 \pi \epsilon_{0} r_{1}}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$V=\frac{Q}{4 \pi \in_{0}}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right]$

$=\frac{Q\left(r_{1}-r_{2}\right)}{4 \pi \in_{0} r_{1} r_{2}}$

Capacitance of the given system is given by,

$C=\frac{\text { Charge }(Q)}{\text { Potential difference }(V)}$

$=\frac{4 \pi \in_{0} r_{1} r_{2}}{r_{1}-r_{2}}$

Hence, proved.