Question

There are four concentric shells $\text{A, B, C}$ and $\text{D}$ of radii of $a,2a,3a$ and $4a$ respectively. Shells $\text{B}$ and $\text{D}$ are given charges $+q$ and $-q$ , respectively. Shell $\text{C}$ is now earthed. The potential difference $V_A-V_C\text{is}\frac{kq}{na}$. Then find the value of $n$.

Answer 1

Let shell $\text{C}$ acquires charge $q{}^{\prime }$ which will be such that potential of $\text{C}$ is zero.


$V_C=\frac{kq}{3a}+\frac{k{q}^{\prime }}{3a}+\left(\frac{-kq}{4a}\right)=0$

$\frac{kq}{3a}+\frac{k{q}^{\prime }}{3a}=\frac{kq}{4a}$

$\Rightarrow q^{\prime }=3q(\frac{1}{4}-\frac{1}{3})$

$\therefore q^{\prime }=-\frac{q}{4}$

As $V_C=0$

$V_A-V_C=V_A$

Now calculating $V_A,$ we get,

$V_A=\frac{kq}{2a}-\frac{k\left(\frac{q}{4}\right)}{3a}-\frac{kq}{4a}$

$\Rightarrow V_A=\frac{kq}{6a}$

$V_A-V_C=\frac{kq}{6a}\text{(or)}$

$\therefore n=6$