Question

Two charged spheres having radii 'a' and 'b' are joined with a wire then the ratio of electric field due to its own charge $\frac{E_a}{E_b}$ on their surface is :

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

Answer 1

Answer: (B)

$\frac{b}{a}$

Now, connecting the two conductors makes their potential equal

$\therefore \frac{k{q}_1}{a}=\frac{k{q}_2}{b}\Rightarrow \frac{q_1}{q_2}=\frac{a}{b}$

$\frac{E_a}{E_b}=\frac{\frac{k{q}_1}{a^2}}{\frac{k{q}_2}{b^2}}$

$=\frac{b^2}{a^2}\times \frac{a}{b}$

$=\frac{b}{a}$