There are four concentric shells A, B, C and | vedclass

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(D) Let the charge on shell $C$ be $q'$.Since shell $C$ is earthed,its potential $V_C = 0$.The potential at shell $C$ (radius $3a$) is due to charges

Vedclass · Accepted Answer

$\frac{Kq}{6a}$

Vedclass · Answer

(D) Let the charge on shell $C$ be $q'$.Since shell $C$ is earthed,its potential $V_C = 0$.The potential at shell $C$ (radius $3a$) is due to charges on $B, C,$ and $D$:$V_C = \frac{Kq}{3a} + \frac{Kq'}{3a} - \frac{Kq}{4a} = 0$$\frac{q}{3a} + \frac{q'}{3a} = \frac{q}{4a}$$q + q' = \frac{3q}{4} \implies q' = -\frac{q}{4}$.Now,calculate the potential at shell $A$ (radius $a$):$V_A = \frac{Kq}{2a} + \frac{Kq'}{3a} - \frac{Kq}{4a}$Substitute $q' = -\frac{q}{4}$:$V_A = \frac{Kq}{2a} + \frac{K(-q/4)}{3a} - \frac{Kq}{4a} = \frac{Kq}{2a} - \frac{Kq}{12a} - \frac{Kq}{4a}$$V_A = Kq \left( \frac{6 - 1 - 3}{12a} \right) = \frac{2Kq}{12a} = \frac{Kq}{6a}$.Since $V_C = 0$,the potential difference $V_A - V_C = \frac{Kq}{6a} - 0 = \frac{Kq}{6a}$.

There are four concentric shells $A, B, C$ and $D$ of radii $a, 2a, 3a$ and $4a$ respectively. Shells $B$ and $D$ are given charges $+q$ and $-q$ respectively. Shell $C$ is now earthed. The potential difference $V_A - V_C$ is:

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