Question

There are two concentric spherical shells of radii $r$ and $2r$. Initially, a charge $Q$ is given to the inner shell. Initially both $S_1$ and $S_2$ are open. Now, switch ${\text{S}}_1$ is closed keeping ${\text{S}}_2$ opened and after a moment ${\text{S}}_2$ is closed and ${\text{S}}_1$ is opened and this process is repeated $n$ times for both the keys alternatively. Find the final potential difference between the shells.

• A. $\frac{KQ}{2^{n}r}$
• B. $\frac{KQ}{n^{2}r}$
• C. $\frac{KQ}{2^{(n+1)}r}$
• D. $\frac{KQ}{r^{(n+1)}}$

Answer 1

The correct option is C $\frac{KQ}{2^{(n+1)}r}$

When we connect the spherical shell to ground the charge on the grounded shell will organize such that the potential of the shell is equal to zero.

Let's assume that after closing ${\text{S}}_2$ the charge on the outer shell be $q_1$.

Hence, the potential at the surface of the outer sphere due to the outer sphere charge is given by equation ,

$V_1=-\frac{k{q}_1}{2r}$

Now, the potential at the surface of the outer sphere due to the inner is given by equation,

$V_2=-\frac{k{Q}^{}}{2r}$

Hence, the total potential is,
$V^{}=V_1+V_2=-(\frac{k{q}_1}{2r}+\frac{kQ}{2r})$

This potential should be zero, as it is connected to the ground.

Hence, we can write,

$0=-(\frac{kQ}{2r}+\frac{k{q}_1}{2r})$

$\Rightarrow q^{}=-Q$

Now, when we close ${\text{S}}_{1,}$ let's assume the charge at the inner sphere is, $q_1^{\prime }$.

The potential due to the outer sphere at the surface of the inner sphere is given by ,

$V_1^{\prime }=-\frac{k(-Q)}{2r}=\frac{kQ}{2r}$

The potential due to the inner sphere at the same location is given by,

$V_2^{\prime }=-\frac{k{q}_1^{\prime }}{r}$

Similarly we can write that,

$V^{\prime }=V_1^{\prime }+V_2^{\prime }=\frac{kQ}{2r}-\frac{k{q}_1^{\prime }}{r}$

$\Rightarrow 0=\frac{kQ}{2r}-\frac{k{q}_1^{\prime }}{r}$

$\Rightarrow q_1^{\prime }=\frac{Q}{2}$

Now, again after closing ${\text{S}}_2$ we can write,

$\frac{k\left(\frac{Q}{2}\right)}{2r}+\frac{k{q}_2}{2r}=0$

$\Rightarrow q_2=-\frac{Q}{2}$

After closing ${\text{S}}_1$ we can write,

$\frac{kQ}{2\times 2r}-\frac{k{q}_2^{\prime }}{r}=0$

$\Rightarrow q_2^{\prime }=\frac{Q}{4}$

So, this follows a geometric progression.

Hence, after $n$ times, the charge will be,
$q_n=-\frac{Q}{2^{n-1}}$

$q_n^{\prime }=\frac{Q}{2^n}$

Hence the potential difference will be,

$V_n=k{q}_n^{\prime }(\frac{1}{r}-\frac{1}{2r})$

On substituting the value of $q_n^{\prime }$ we get,

$V_n=\frac{kQ}{2^n}(\frac{1}{r}-\frac{1}{2r})$

$\therefore V_n=\frac{kQ}{2^{n+1}r}$

Hence, option (c) is correct answer.

Why this question : A highly conceptual question that can make the students revise the entire concept of earthing and switching at once.