Solution
Let the radius of the shell be $R$. The shell is earthed, so its potential is zero ($V_{shell} = 0$). Let the final charge on the shell be $Q_{final}$.
The potential at the surface of the shell is the scalar sum of potentials due to all charges:
- Potential due to inner charge $q$ at distance $0.5R$ (which is effectively at the center for external points, but potential is constant $k q / R$ everywhere on surface): $V_1 = \frac{kq}{R}$.
- Potential due to outer charge $2q$ at distance $2R$: $V_2 = \frac{k(2q)}{2R} = \frac{kq}{R}$.
- Potential due to the charge on the shell itself ($Q_{final}$): $V_{shell} = \frac{kQ_{final}}{R}$.
Condition for Earthing:
$$V_{net} = V_1 + V_2 + V_{shell} = 0$$ $$\frac{kq}{R} + \frac{kq}{R} + \frac{kQ_{final}}{R} = 0$$ $$q + q + Q_{final} = 0$$ $$Q_{final} = -2q$$Charge Flow:
The initial charge on the shell was $q$. The final charge is $-2q$.
The charge that flows from the shell to the earth is $q – (-2q) = 3q$.
Since the question asks “how much charge will flow to the earth”, the answer is the amount lost by the system.
Answer: (b) 3q