Physics - Electrostatics Question with Solution | TestHub

Initially, the inner sphere has charge $\mathrm{Q}$ and the outer shell is uncharged. When the outer shell is momentarily grounded, its potential becomes zero. The potential of the outer shell is ${\mathrm{V}}_{\mathrm{outer}}=\mathrm{k}\frac{\mathrm{Q}}{2R}+\mathrm{k}\frac{{\mathrm{Q}}_{\mathrm{shell}}}{2R}$. Setting ${\mathrm{V}}_{\mathrm{outer}}=0$ gives ${\mathrm{Q}}_{\mathrm{shell}}=-\mathrm{Q}$. So, the outer shell acquires a charge of $-\mathrm{Q}$ from the ground. After grounding, the system consists of the inner sphere with charge $\mathrm{Q}$ and the outer shell with charge $-\mathrm{Q}$. When the inner sphere is connected to the outer shell by a conducting wire, they form a single conductor. All the charge in a connected system of conductors resides on the outermost surface. The total charge of the system is $\mathrm{Q}+(-\mathrm{Q})=0$. Since the total charge is zero and resides on the outer surface of the outer shell, the final charge on the inner solid sphere must be zero.