The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is

Q: The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is

(a) F/4

(b) 4F

(c) 16F

(d) F

Ans: (d)

Sol: $\large F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$  …..(i)

In medium ,

$\large F’ = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{(r/2)^2} $  …..(ii)

On dividing ,

$\large \frac{F’}{F} = \frac{4}{\epsilon_r} = \frac{4}{4}= 1$

F’ = F