Question

Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of dielectric constant $4$ is introduced between them, the force will be:

• A. $4F$
• B. $2F$
• C. $F/2$
• D. $F/4$

Answer 1

The correct option is D $F/4$

$\text{Step 1: Force between two charges in vaccum}$

Applying Coulomb's law,

$F=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{d^2}$ $....\left(1\right)$

where $d$ is the distance between the charges

$\text{Step 2: Force between charges in medium}$

Ler $k$ be the dielectric constant of the medium.

$F^{\prime }=\frac{1}{4\pi {\in }_0}\frac{1}{k}\frac{q_1{q}_2}{d^2}$

$\Rightarrow F^{\prime }=\frac{F}{k}=\frac{F}{4}$

Hence, option (D) is correct.