Question

Two concentric shells of radii R and 2R are shown in Fig. 3.115.Initially, a charge q is imparted to the inner shells. Now, key $K_1$ is closed and opened and then key $K_2$ is closed and opened. After the keys $K_1$ and $K_2$ are alternately closed n times each, find the potential difference between the shells. Note that finally key $K_2$ remains closed.

Answer 1

When $K_1$ is closed first time, the outer sphere is earthed and the potential on it becomes zero. Let the charge on it be $q_1^{{}^{\prime }}$ potential due to charge on the inner sphere and that due to charge on the outer sphere is

$V_1^{{}^{\prime }}=\frac{1}{4\pi {\epsilon }_0}[\frac{q}{2R}+\frac{q_1^{{}^{\prime }}}{2R}]=0$ or $q_1^{{}^{\prime }}=-q$

When $K_2$ is closed first time, the potential $V_2^{{}^{\prime }}$ on the inner sphere becomes zero as it is earthed. Let the new charge on inner sphere be $q_2^{{}^{\prime }}$

$0=\frac{1}{4\pi {\epsilon }_0}\frac{q_2^{{}^{\prime }}}{R}+\frac{1}{4\pi {\epsilon }_0}\frac{(-q)}{\left(2R\right)}or{q}_2^{{}^{\prime }}=\frac{q}{2}$

Now, when $K_1$ will be closed second time, charge on the outer sphere will be $-q_2^{{}^{\prime }}$, i.e., - q/2. After one event involving closure and opening of $K_1$ and $K_2$, charge is reduced to half of its initial value. Similarly, when $K_1$ will be closed $n^{th}$ time,charge on the outer sphere will be $-q/(2^{n-1}$ as each time charge will be reduced to half of the previous value.

After closing $K_2$ $n^{th}$ time, charge on the inner shell will be negative of half the charge on the outer shell,i.e., (+q2") and potential on it will be zero. For potential of the outer shell

$V_0=\frac{1}{4\pi {\epsilon }_0}\frac{(+q/2^n)}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{(-q/2^{n-1})}{2R}$

$\frac{-q[-1+2]}{4\pi {\epsilon }_0{2}^{n+1}R}=\frac{-q}{4\pi {\epsilon }_0{2}^{n+1}R}$

potential difference is

$V_0-V_1=\frac{-q}{4\pi {\epsilon }_0{2}^{n+1}R}-0=\frac{-q}{4\pi {\epsilon }_0{2}^{n+1}R}$