An empty thick conducting shell of inner radi | vedclass

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Question

(A) When a conducting shell is earthed,its potential becomes zero. Since the potential of the entire conductor (including the inner and outer surfaces

Vedclass · Accepted Answer

Only the potential of the outer surface becomes zero.

Vedclass · Answer

(A) When a conducting shell is earthed,its potential becomes zero. Since the potential of the entire conductor (including the inner and outer surfaces) must be the same,the potential of the whole shell becomes zero. Let $q_{in} = -4\pi a^2 \sigma$ be the charge on the inner surface and $q_{out} = 4\pi b^2 \sigma'$ be the charge on the outer surface. The potential at the surface of the shell is given by $V = \frac{1}{4\pi \epsilon_0} \left( \frac{q_{in}}{b} + \frac{q_{out}}{b} ight) = 0$. This implies $q_{in} + q_{out} = 0$,so $q_{out} = -q_{in}$. Since $q_{in}$ is negative,$q_{out}$ must be positive. However,if the shell was initially charged such that $q_{out} 
eq -q_{in}$,earthing it causes charge to flow from the earth to the outer surface until the total potential becomes zero. Thus,the charge on the outer surface becomes $q_{out} = -q_{in} = 4\pi a^2 \sigma$. Since the potential of the entire conductor is zero,the charge on the outer surface is no longer determined by the initial $\sigma'$,but by the requirement that the total potential is zero. Therefore,the charge on the outer surface becomes $4\pi a^2 \sigma$,which is positive. Thus,the potential of the entire shell becomes zero,and the charge on the outer surface changes to neutralize the effect of the inner charge.

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