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

Answer 1

The correct option is A $\frac{4\pi {\epsilon }_0{r}_1{r}_2}{r_1-r_2}$

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

$V=\frac{q}{4\pi {\epsilon }_0{r}_2}-\frac{q}{4\pi {\epsilon }_0{r}_1}=\frac{q}{4\pi {\epsilon }_0}[\frac{1}{r_2}-\frac{1}{r_1}]$

$V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{r_1-r_2}{r_1{r}_2}\right)....\left(i\right)$

Also $C=\frac{q}{V}$

$\therefore C=\frac{4\pi {\epsilon }_0{r}_1{r}_2}{r_1-r_2}$ Using (i)