Question

The figure shows a spherical capacitor with inner sphere earthed. If $a=2\text{cm}$ and $b=3\text{cm}$, then the capacitance of the system is
(Take $k=9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$)

• A. $40\text{pF}$
• B. $20\text{pF}$
• C. $5\text{pF}$
• D. $10\text{pF}$

Answer 1

The correct option is D $10\text{pF}$


Let $q$ charge is given to outer sphere and charge appearing on inner sphere is $q^{\prime }$.

Radii of the two spherical conductors are $a$ and $b$ with $b>a$.

Since, inner sphere is earthed, so

$V_a=0$

$\Rightarrow \frac{k{q}^{\prime }}{a}+\frac{kq}{b}=0$

$\Rightarrow q^{\prime }=-\frac{a}{b}q$

Potential difference, $\Delta V=V_b-V_a$

$\Delta V=V_b(\because V_a=0)$

where, $V_b=\frac{k{q}^{\prime }}{b}+\frac{kq}{b}$

$\Rightarrow V_b=-\frac{kqa}{b^2}+\frac{kq}{b}$

$\Rightarrow V_b=\frac{kq(b-a)}{b^2}$

$\therefore \Delta V=V_b=\frac{q(b-a)}{4\pi {\epsilon }_0{b}^2}$

Using $q=CV$, we get

$C=\frac{4\pi {ϵ}_0{b}^2}{b-a}$


Capacitance of the spherical capacitor in this case,

$C=\frac{4\pi {\epsilon }_0{b}^2}{(b-a)}$

Given: $b=3\times {10}^{-2}\text{m}\&a=2\times {10}^{-2}\text{m}$

$\Rightarrow C=\frac{1}{9\times {10}^9}\times \frac{9\times {10}^{-4}}{{10}^{-2}}$

$\therefore C={10}^{-11}=10\text{pF}$

Key concept: Capacitance of spherical capacitor with inner conductor earthed.