A metal sphere of radius r carrying charge q | vedclass

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(C) The potential at the surface of the inner sphere of radius $r$ due to its own charge $q$ is $V_{q,r} = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.The p

Vedclass · Accepted Answer

$V = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{R} \right)$

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(C) The potential at the surface of the inner sphere of radius $r$ due to its own charge $q$ is $V_{q,r} = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.The potential at the surface of the inner sphere due to the charge $Q$ on the outer shell of radius $R$ is $V_{Q,r} = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$ (since the point is inside the shell).Total potential of the inner sphere is $V_{inner} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right)$.The potential at the surface of the outer shell of radius $R$ due to the charge $q$ is $V_{q,R} = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.The potential at the surface of the outer shell due to its own charge $Q$ is $V_{Q,R} = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.Total potential of the outer shell is $V_{outer} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{R} + \frac{Q}{R} \right)$.The potential difference $\Delta V = V_{inner} - V_{outer} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} - \frac{q}{R} - \frac{Q}{R} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{R} \right)$.

$A$ metal sphere of radius $r$ carrying charge $q$ is placed at the center of a metallic shell of radius $R$ carrying charge $Q$. Find the expression for the potential difference between the sphere and the shell.

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