Q74 / 90 · JEE Main 2024 (31 Jan Shift 2) · Physics

Force between two point charges $q_{1}$ and $q_{2}$ placed in vacuum at $r$ $cm$ apart is $F$ . Force between them when placed in a medium having dielectric $K = 5$ at $\frac{r}{5}$ $cm$ apart will be:

Solution

Using Coulomb's law, we can write:

In air $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (K ε_{0})} \frac{q_{1} q_{2}}{{(r')}^{2}} = \frac{1}{4 π (5 ε_{0})} \frac{25 q_{1} q_{2}}{{(r)}^{2}} = 5 F$

Hence, the correct answer is option: 'B'.