Question

Two charged conducting spheres \(S_1\) and \(S_2\) of radii \(8\) cm and \(18\) cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on \(S_1\) and \(S_2\) spheres are \(E_{S_1}\) and \(E_{S_2}\) respectively. The value of \( \dfrac{E_{S_1}}{E_{S_2}} \) is:

• A. \( \dfrac{3}{2} \)
• B. \( \dfrac{2}{3} \)
• C. \( \dfrac{4}{9} \)
• D. \( \dfrac{9}{4} \)

Answer 1

The Correct Option is D

Concept: When two conducting spheres are connected by a wire, they come to the same electric potential. For a charged conducting sphere: \[ V=\frac{kQ}{R} \] and the electric field at the surface: \[ E=\frac{kQ}{R^2} \]
Step 1:Use equal potential condition} \[ \frac{kQ_1}{R_1}=\frac{kQ_2}{R_2} \] \[ \frac{Q_1}{Q_2}=\frac{R_1}{R_2} \] \[ \frac{Q_1}{Q_2}=\frac{8}{18}=\frac{4}{9} \]
Step 2:Find the ratio of electric fields} \[ E_1=\frac{kQ_1}{R_1^2}, \qquad E_2=\frac{kQ_2}{R_2^2} \] \[ \frac{E_1}{E_2} = \frac{Q_1}{Q_2} \cdot \frac{R_2^2}{R_1^2} \] Substitute values: \[ \frac{E_1}{E_2} = \frac{4}{9} \cdot \frac{18^2}{8^2} \] \[ = \frac{4}{9} \cdot \frac{324}{64} \] \[ = \frac{1296}{576} \] \[ =\frac{9}{4} \] Thus \[ \boxed{\frac{9}{4}} \]