# 2.29        A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports figure. Show that the capacitance of a spherical capacitor is given by $C = \frac{4\pi \epsilon_{0}r_{1}r_{2}}{r_{1}-r_{2}}$ where r1 and r2 are the radii of outer and inner spheres, respectively.

Given

the radius of the outer shell = $r_1$

the radius of the inner shell = $r_2$

charge on Inner surface of outer shell =  $Q$

Induced charge on the outer surface of inner shell = $-Q$

NOW,

The potential difference between the two shells

$V=\frac{Q}{4\pi \epsilon _0r_2}-\frac{Q}{4\pi \epsilon _0r_1}$

Now Capacitance is given by

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

$C=\frac{Q}{\frac{Q(r_1-r_2)}{4\pi \epsilon _0r_1r_2}}=\frac{4\pi \epsilon _0r_1r_2}{r_1-r_2}$

Hence proved.

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