A solid conducting sphere, having a charge Q, | vedclass

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(D) Let the radius of the solid sphere be $r_1$ and the radius of the hollow shell be $r_2$.In the first condition, the solid sphere has charge $Q$ an

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$1$

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(D) Let the radius of the solid sphere be $r_1$ and the radius of the hollow shell be $r_2$.In the first condition, the solid sphere has charge $Q$ and the shell is uncharged.The potential of the solid sphere is $V_{in} = \frac{kQ}{r_1} + \frac{k(0)}{r_2} = \frac{kQ}{r_1}$.The potential of the hollow shell is $V_{out} = \frac{kQ}{r_2} + \frac{k(0)}{r_2} = \frac{kQ}{r_2}$.The potential difference is $V = V_{in} - V_{out} = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.In the second condition, the shell is given a charge of $-4Q$.The potential of the solid sphere is $V'_{in} = \frac{kQ}{r_1} + \frac{k(-4Q)}{r_2} = \frac{kQ}{r_1} - \frac{4kQ}{r_2}$.The potential of the hollow shell is $V'_{out} = \frac{kQ}{r_2} + \frac{k(-4Q)}{r_2} = -\frac{3kQ}{r_2}$.The new potential difference is $V' = V'_{in} - V'_{out} = \left( \frac{kQ}{r_1} - \frac{4kQ}{r_2} \right) - \left( -\frac{3kQ}{r_2} \right)$.$V' = \frac{kQ}{r_1} - \frac{4kQ}{r_2} + \frac{3kQ}{r_2} = \frac{kQ}{r_1} - \frac{kQ}{r_2} = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = V$.Thus, the potential difference remains $V$.

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