Q 12-01-066JEE MainJEE Main 2024 (31 Jan, Shift 2)Easy
The force between two point charges $q_1$ and $q_2$ placed in vacuum at $r$ cm apart is $F$. The force between them when placed in a medium having dielectric constant $K = 5$ at $\dfrac r5$ cm apart will be:
Answer: (B) $5F$
$$F' = \frac{1}{K}\cdot\frac{kq_1q_2}{(r/5)^2} = \frac{25}{5}F = 5F$$
Solution by Sreeraj P, M.Sc Physics