A metal sphere of radius R carrying charge q | vedclass

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(C) $1$. The inner sphere has charge $q$. Due to induction, a charge $-q$ is induced on the inner surface of the shell (radius $a$).$2$. Since the she

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$\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{R}-\frac{1}{a}\right)$

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(C) $1$. The inner sphere has charge $q$. Due to induction, a charge $-q$ is induced on the inner surface of the shell (radius $a$).$2$. Since the shell is grounded, its potential is zero. Let the charge on the outer surface (radius $b$) be $q'$. The total charge on the shell is $Q_{shell} = -q + q' = 0$, which implies $q' = q$. However, because the shell is grounded, the charge $q'$ flows to the ground, leaving the outer surface with zero charge. Thus, the inner surface of the shell has charge $-q$ and the outer surface has charge $0$.$3$. The potential at the centre of the sphere is the sum of potentials due to the sphere (radius $R$, charge $q$), the inner surface of the shell (radius $a$, charge $-q$), and the outer surface of the shell (radius $b$, charge $0$).$4$. $V_{centre} = V_{sphere} + V_{inner_shell} + V_{outer_shell} = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R} + \frac{1}{4 \pi \epsilon_{0}} \frac{-q}{a} + \frac{1}{4 \pi \epsilon_{0}} \frac{0}{b} = \frac{q}{4 \pi \epsilon_{0}} \left( \frac{1}{R} - \frac{1}{a} \right)$.

$A$ metal sphere of radius $R$ carrying charge $q$ is surrounded by a thick concentric metal shell of inner and outer radii $a$ and $b$ respectively. The net charge on the shell is zero. The potential at the centre of the sphere, when the outer surface of the shell is grounded, will be

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