An empty thick conducting shell of inner radi | vedclass

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Question

(C) For a conducting shell, the electric field inside the material of the conductor is zero. Since $E = -\frac{dV}{dr} = 0$, the potential $V$ must be

Vedclass · Accepted Answer

Both the surfaces have equal potential.

Vedclass · Answer

(C) For a conducting shell, the electric field inside the material of the conductor is zero. Since $E = -\frac{dV}{dr} = 0$, the potential $V$ must be constant throughout the volume of the conductor. Therefore, the potential at the inner surface (radius $a$) and the outer surface (radius $b$) must be equal. Let's calculate the potential $V$ at any point on the shell. The potential due to the inner surface charge $q_{in} = -\sigma(4\pi a^2)$ and outer surface charge $q_{out} = \sigma(4\pi b^2)$ at a distance $r$ (where $a \le r \le b$) is: $V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_{in}}{r} + \frac{q_{out}}{b} \right]$ $V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{-\sigma(4\pi a^2)}{r} + \frac{\sigma(4\pi b^2)}{b} \right] = \frac{\sigma}{\varepsilon_0} (b - \frac{a^2}{r})$. At the outer surface $(r=b)$, $V = \frac{\sigma}{\varepsilon_0} (b - \frac{a^2}{b}) = \frac{\sigma}{\varepsilon_0 b} (b^2 - a^2)$. However, in the context of a conductor, the potential is uniform. Since the question asks for the correct statement, and the potential of a conductor is constant, both surfaces have equal potential. Thus, option $C$ is correct.

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