Question

Two thin conducting shells of radii $R$ and $3R$ are shown in the figure. The outer shell carries a charge $+Q$ and the inner shell is neutral. The inner shell is earthed with the help of a switch $S$.

• A. With the switch $S$ closed, the charge attained by the inner sphere is $-1/3$ times that of charge present on the outer sphere.
• B. By closing the switch the capacitance of the system increases
• C. With the switch $S$ open, the potential of the inner sphere is equal to that of the outer
• D. When the switch $S$ is closed, the potential of the inner sphere becomes zero

Answer 1

Answer: (A), (B), (C), (D)

When the switch $S$ is closed, the potential of the inner sphere becomes zero

When the switch $S$ is open, no charge is present on inner shell.

$V_{\text{inner shell}}=V_{\text{outer shell}}=\frac{kQ}{3R}$

So, Option $\left(a\right)$ is correct

Let the charge on the inner shell after closing the switch be $Q^{\prime }$


Since the inner shell is earthed, the net potential must be zero for it.
$V_{\text{inner}}=0$

The net potential at the inner shell is the sum of potential due to the inner shell itself and due to the outer shell.

$\Rightarrow \frac{k{Q}^{\prime }}{R}+\frac{kQ}{3R}=0\Rightarrow Q^{\prime }=-\frac{Q}{3}$

So, Option $\left(b\right)$ and $\left(c\right)$ are correct.

Now from the definition of capacitance,
$C_i=4\pi {\epsilon }_0\left(3R\right)$

The charge on inner shell is zero initially when switch is open, hence only capacitance of outer sphere is taken into account.

$\&C_f=4\pi {\epsilon }_0\left(3R\right)+\frac{4\pi {ϵ}_0}{(\frac{1}{R}-\frac{1}{3R})}$

After the switch is closed the capacitance of both spheres add up, since they get connected in parallel.

$\Rightarrow C_f>C_i$

Hence, options (a) , (b) , (c) and (d) are the correct answers.