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(D) When a conducting sphere is earthed, its potential becomes $0$.Let the charge on the inner surface of the hollow sphere be $q_{in}$ and the charge

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Zero

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(D) When a conducting sphere is earthed, its potential becomes $0$.Let the charge on the inner surface of the hollow sphere be $q_{in}$ and the charge on the outer surface be $q_{out}$.By the property of induction, $q_{in} = -q$.Since the outer sphere is earthed, its potential $V$ is the sum of potentials due to the inner sphere (charge $q$) and the hollow sphere (charges $q_{in}$ and $q_{out}$).The potential at the surface of the outer sphere (radius $2R$) is given by:$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{2R} + \frac{q_{in}}{2R} + \frac{q_{out}}{2R} \right) = 0$.Substituting $q_{in} = -q$:$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{2R} - \frac{q}{2R} + \frac{q_{out}}{2R} \right) = 0$.This simplifies to $\frac{q_{out}}{2R} = 0$, which implies $q_{out} = 0$.Therefore, the charge on the outer surface is $0$.

$A$ charge $q$ is distributed uniformly on the surface of a sphere of radius $R$. It is covered by a concentric hollow conducting sphere of radius $2R$. What will be the charge on the outer surface of the hollow sphere if it is earthed?

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