Question

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports as shown in figure.

The capacitance $C$, of this spherical capacitor is

• A. $\frac{4\pi {\epsilon }_0{r}_1{r}_2}{r_1-r_2}$
• B. $\frac{4\pi {\epsilon }_0\left(r_2-r_1\right)}{r_1{r}_2}$
• C. $\frac{r_1{r}_2}{4\pi {\epsilon }_0\left(r_2-r_1\right)}$
• D. $\frac{\left(r_1-r_2\right)}{4\pi {\epsilon }_0{r}_1{r}_2}$

Answer 1

Answer: (A)

As shown in figure, $+q$ charge spreads uniformly on inner surface of outer sphere of radius $r_1$. The induced charge $-q$ spreads uniformly on the outer surface of inner sphere of radius $r_2$.
The outer surface of outer sphere is earthed. Due to electrostatic shielding
$E=0$ for $r<r_2$ and $E=0$ for $r>r_1$
In the space between the two spheres, Potential difference between two spheres,
$V=\frac{q}{4\pi {\epsilon }_0{r}_2}-\frac{q}{4\pi {\epsilon }_0{r}_1}$
$=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{r_2}-\frac{1}{r_1}\right]$
$V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{r_1-r_2}{r_1{r}_2}\right)\dots \left(i\right)$
Also $C=\frac{q}{V}$
$\therefore C=\frac{4\pi {\epsilon }_0{r}_1{r}_2}{r_1-r_2}$ (using $\left(i\right))$