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(D) Let the radius of the inner sphere be $a$ and the radius of the outer shell be $b$.Initially,the potential of the sphere is $V_{	ext{sphere}} = \

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$V$

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(D) Let the radius of the inner sphere be $a$ and the radius of the outer shell be $b$.Initially,the potential of the sphere is $V_{	ext{sphere}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{a}$ and the potential of the shell is $V_{	ext{shell}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{b}$.The potential difference is $V = V_{	ext{sphere}} - V_{	ext{shell}} = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} ight)$.Now,a charge of $-3Q$ is given to the shell.The new potential of the sphere is $V'_{	ext{sphere}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} ight)$.The new potential of the shell is $V'_{	ext{shell}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{b} - \frac{3Q}{b} ight) = \frac{1}{4\pi\epsilon_0} \left( -\frac{2Q}{b} ight)$.The new potential difference is $V' = V'_{	ext{sphere}} - V'_{	ext{shell}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} - (-\frac{2Q}{b}) ight) = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} + \frac{2Q}{b} ight) = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} ight) = V$.Thus,the potential difference remains $V$.

$A$ sphere with charge $Q$ is placed concentrically inside an uncharged spherical shell. The potential difference between them is $V$. Now,if a charge of $-3Q$ is given to the spherical shell,what will be the potential difference between them?

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