Question

Two concentric shells of radii $R$ and $2R$ are shown in the figure. Initially a charge $q$ is imparted to the inner shell. After the keys $K_1$ & $K_2$ are alternately closed $n$ times each, find the potential difference between the shells.

Answer 1

When $K_1$ is closed first time, outer sphere is earthed and its potential becomes zero. Let the charge on it be $q_1^{\prime }$.
$V_1^{\prime }=$ Potential due to charge on inner sphere and that due to charge on outer sphere.
$0=\frac{1}{4\pi {\epsilon }_0}[\frac{q}{2R}+\frac{q_1}{2R}]$
$⟹q_1^{\prime }=-q$
When $K_2$ is closed first time, the potential $V_2^{\prime }$ on inner sphere becomes zero as it is earthed.
Let the new charge on inner sphere be $q_{2^{\prime }}$
$0=\frac{1}{4\pi {\epsilon }_0}\frac{q_2^{\prime }}{R}+\frac{1}{4\pi {\epsilon }_0}\frac{(-q)}{\left(2R\right)}$
$⟹q_2^{\prime }=q/2$
Now when $K-1$ will be closed second time charge on under sphere will be $-q_2^{\prime }$ i.e. $-q/2$
Similarly when $K_1$ will be closed $n^{th}$ time, charge on outer sphere will be $-\frac{q}{2^{n-1}}$ as each time.
Charge will be reduced to half the previous value.
After closing $K_2$ $n^{th}$ time charge on inner shell will be negative of half the charge on outer shell.
i.e. $(-\frac{q}{2^n})$ and potential on it will be zero.
For potential of outer shell
$V_0=\frac{1}{4\pi {\epsilon }_0}\frac{(-q/2^n)}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{\left(\frac{q}{2^{n-1}}\right)}{2R}$
$V_0=\frac{q[-1+2]}{4\pi {\epsilon }_0{2}^{n+1}R}=+\frac{q}{4\pi {\epsilon }_0{2}^{n+1}R}-0=\frac{q}{4\pi {\epsilon }_0{2}^{n+1}R}$