Q 12-01-074JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
If two charges $q_1$ and $q_2$ are separated with distance $d$ and placed in a medium of dielectric constant $k$, what will be the equivalent distance between charges in air for the same electrostatic force?
Answer: (A) $d\sqrt k$
$$\frac{q_1q_2}{4\pi\varepsilon_0kd^2}=\frac{q_1q_2}{4\pi\varepsilon_0r^2}\ \Rightarrow\ r=d\sqrt k$$
Solution by Sreeraj P, M.Sc Physics