Q 12-01-036JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
A small uncharged conducting sphere is placed in contact with an identical sphere but having $4\times10^{-8}\ \text{C}$ charge and then removed to a distance such that the force of repulsion between them is $9\times10^{-3}\ \text{N}$. The distance between them is (Take $\dfrac{1}{4\pi\epsilon_0}$ as $9\times10^9$ in SI units)
Answer: (B) $2\ \text{cm}$
Identical spheres share the charge equally: $q = 2\times10^{-8}\ \text{C}$ each.
$$F = \frac{kq^2}{r^2} \Rightarrow r^2 = \frac{9\times10^9\times(2\times10^{-8})^2}{9\times10^{-3}} = \frac{3.6\times10^{-6}}{9\times10^{-3}} = 4\times10^{-4}\ \text{m}^2$$
$r = 0.02\ \text{m} = 2\ \text{cm}$
Solution by Sreeraj P, M.Sc Physics