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. $\frac{4\pi\varepsilon_{0} r_{1}r_{2}}{r_{1}-r_{2}}$
• B. $\frac{4\pi\varepsilon_{0} \left(r_{2}-r_{1}\right)}{r_{1}r_{2}}$
• C. $\frac{r_{1}r_{2}}{4 \pi\varepsilon_{0} \left(r_{2}-r_{1}\right)}$
• D. $\frac{\left(r_{1}-r_{2}\right)}{4\pi\varepsilon_{0}r_{1}r_{2}}$

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,
$V=\frac{q}{4\pi\varepsilon_{0} r_{2}}-\frac{q}{4\pi\varepsilon_{0} r_{1}}$
$=\frac{q}{4\pi\varepsilon_{0}} \left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right]$
$V=\frac{q}{4\pi\varepsilon_{0}} \left(\frac{r_{1}-r_{2}}{r_{1}r_{2}}\right) \ldots\left(i\right)$
Also $C= \frac{q}{V}$
$\therefore C=\frac{4\pi\varepsilon_{0} r_{1}r_{2}}{r_{1}-r_{2}}$ (using $\left(i\right))$