Question

There are four concentric shells $\text{A, B, C}$ and $\text{D}$ of radii $a,2a,3a$ and $4a$ respectively. Shells $\text{B}$ and $\text{D}$ are given charges $+q$ and $-q$ respectively. Shell $\text{C}$ is now earthed. Find the potential difference $V_A-V_C$ (Where $K=\frac{1}{4\pi {ϵ}_0}$)

• A. $\frac{Kq}{3a}$
• B. $\frac{2Kq}{3a}$
• C. $\frac{Kq}{6a}$
• D. $\frac{3Kq}{2a}$

Answer 1

The correct option is C $\frac{Kq}{6a}$


Let charge on the shell $\text{C}$ be $q^{\prime }$.

Since the shell $\text{C}$ is earthed its potential will be zero.i.e., $V_C=0...\left(1\right)$

From figure potential of $\text{C}$ can be written as,

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

$\Rightarrow \frac{Kq}{3a}+\frac{K{q}^{{}^{\prime }}}{3a}=+\frac{Kq}{4a}$

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

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

Now, potential of shell $\text{A}$,

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

$\Rightarrow V_A=\frac{Kq[6-1-3]}{12a}$

$\Rightarrow V_A=\frac{Kq}{6a}...\left(2\right)$

Therefore,
$V_A-V_C=\frac{Kq}{6a}-0$

From $\left(1\right)$ and $\left(2\right)$,

$\therefore V_A-V_C=\frac{Kq}{6a}$

Hence option (c) is correct answer.