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 0 1 2 1 2 4 – r r C r r pe = where r1 and r2 are the radii of outer and inner spheres, respectively.

Answer 1

Let +Q is the charge on the outer spherical shell A of radius $r_1$ and +Q is the charge on inner spherical shell B of radius $r_2$.

Then electric potential on shell A, $V_A=V_{AA}+V_{AB}$

= $\frac{1}{4\pi\epsilon_0}\left[\frac{+Q}{r_1}-\frac{Q}{r_1}\right]=0$

so, $V_A=0$

and electric potential on shell B is $V_B=V_{BB}+V_{BA}$

or, =$\frac{1}{4\pi\epsilon_0}\left[-\frac{Q}{r_2}+\frac{Q}{r_1}\right]$

so, $V_B=\frac{Q}{4\pi\epsilon_0}\left[\frac{r_2-r_1}{r_1r_2}\right]$

so, the potential difference between two spherical shell A and shell B is $V_A-V_B=0-\frac{Q}{4\pi\epsilon_0}\left[\frac{r_2-r_1}{r_1r_2}\right]$

or, V = $\frac{Q}{4\pi\epsilon_0}\left[\frac{r_1-r_2}{r_1r_2}\right]$

we know, Q = CV

so, V = Q/C = $\frac{Q}{4\pi\epsilon_0}\left[\frac{r_1-r_2}{r_1r_2}\right]$

or, C = $\frac{4\pi\epsilon_0r_1r_2}{r_1-r_2}$

hence, the capacitance of the spherical capacitor is $\frac{4\pi\epsilon_0r_1r_2}{r_1-r_2}$