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.
Radius of the outer shell = r1
Radius of the inner shell = r2
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.