Question

Two conducting spheres of radii a and b respectively are charged and joined by a wire. The ratio of electric field due the spheres at their surfaces when isolated is

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{a^2}{b^2}$
• D. $\frac{b^2}{a^2}$

Answer 1

The correct option is B $\frac{b}{a}$

Joined by a wire means they are at the same potential. For same potential $\frac{k{Q}_1}{a_1}=\frac{k{Q}_2}{a_2}\Rightarrow \frac{Q_1}{Q_2}=\frac{a}{b}$
Further, the electric field at the surface of the sphere having radius R and charge Q is $\frac{kQ}{R^2}$
$\therefore \frac{E_1}{E_2}=\frac{\frac{k{Q}_1}{a^2}}{\frac{k{Q}_2}{b^2}}=\frac{Q_1}{Q_2}\times \frac{b^2}{a^2}=\frac{b}{a}$