Question

The figure shows spherical conducting shells with inner sphere earthed. If $a=2\text{cm}$ and $b=3\text{cm}$, then the capacitance of the system is

• 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 be $q^{\prime }$.

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

Since, the inner sphere is earthed, so

$V_a=0$

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

$\Rightarrow q^1=\frac{-a}{b}q$

Now, the potential difference,

$\delta V=V_b-V_a$

$\Rightarrow \delta V=V_b-0=V_b...\left(1\right)$

where,

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

Substituting the value of $q^{\prime }$,

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

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

So, from equation $\left(1\right)$ and $k=\frac{1}{4\pi {\epsilon }_0}$

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

Using, $q=C\delta V$, we get

Capacitance of the system:

$C=\frac{4\pi {ϵ}_0{b}^2}{(b-a)}$

Given,

$b=3\times {10}^{-2}\text{m}$ and $a=2\times {10}^{-2}\text{m}$

$\Rightarrow C=\frac{1}{9\times {10}^9}\times \frac{{(3\times {10}^{-2})}^2}{{10}^{-2}}$

$\Rightarrow C={10}^{-11}\text{F}$

$\therefore C=10\text{pF}$

Hence, option (d) is the correct answer.

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