MHTCET 20 Apr 2026 Shift 2 Paper

Concept:The force between two charges depends on the medium and the distance; introducing a dielectric reduces the force, so distance must be adjusted to restore the original force.Explanation:In air, the force between the charges separated by distance $R$ is given by Coulomb's law:$$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{R^2}$$When the space is filled with a dielectric of constant $K$, the permittivity becomes $K\varepsilon_0$.For the same force $F$ at a new distance $R_1$, we write:$$F = \frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{R_1^2}$$Equating the two expressions for $F$:$$\frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{R^2} = \frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{R_1^2}$$Cancelling common terms gives:$$\frac{1}{R^2} = \frac{1}{K R_1^2}$$Therefore, $K R_1^2 = R^2$, which simplifies to:$$R_1 = \frac{R}{\sqrt{K}}$$Answer:$$R_1 = \frac{R}{\sqrt{K}}$$So the correct option is D.