Question

A spherical metal shell \( A \) of radius \( R_A \) and a solid metal sphere \( B \) of radius \( R_B \) (\( R_B<R_A \)) are kept far apart and each is given charge \( +Q \). If they are connected by a thin metal wire and \( Q_A \) and \( Q_B \) are the charge on \( A \) and \( B \), respectively, then:

• A. \( Q_A = Q_B = 0 \)
• B. \( Q_A = Q_B = Q \)
• C. \( Q_A<Q_B \)
• D. \( Q_A = -Q_B \)
• E. \( Q_A>Q_B \)

Hint:

Charge distribution between connected conductors adjusts to equalize their electric potentials.

Answer 1

Answer: (E)

When a larger spherical shell and a smaller solid sphere are connected, the potential must equalize. Since potential \( V \) for a sphere is given by \( V = \frac{Q}{4\pi \epsilon_0 R} \), and \( R_A>R_B \), the charges rearrange to maintain \( V_A = V_B \), resulting in: \[ \frac{Q_A}{R_A} = \frac{Q_B}{R_B} \implies Q_A R_B = Q_B R_A \] Since \( R_A>R_B \), it follows that \( Q_A>Q_B \) to balance the equation.