A solid conducting sphere having a charge Q i | vedclass

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Question

(A) Let the radius of the solid sphere be $a$ and the radius of the hollow shell be $b$. Initially,the solid sphere has charge $Q$ and the shell has c

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$1$

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(A) Let the radius of the solid sphere be $a$ and the radius of the hollow shell be $b$. Initially,the solid sphere has charge $Q$ and the shell has charge $0$. The potential at the surface of the solid sphere is $V_{	ext{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a}$. The potential at the surface of the hollow shell is $V_{	ext{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b}$. The initial potential difference is $V = V_{	ext{sphere}} - V_{	ext{shell}} = \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} ight)$. Now,the shell is given a charge of $-3Q$. The new potential at the surface of the solid sphere is $V'_{	ext{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b}$. The new potential at the surface of the hollow shell is $V'_{	ext{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{2Q}{b} ight)$. The new potential difference is $V' = V'_{	ext{sphere}} - V'_{	ext{shell}} = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} - (-\frac{2Q}{b}) ight) = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{Q}{b} ight) = V$.

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