Question

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. What will be the distance at which they will experience force $4F$, when placed in a medium of dielectric constant $16$?

• A. $r$
• B. $\frac{r}{4}$
• C. $\frac{r}{8}$
• D. $2r$

Answer 1

The correct option is C $\frac{r}{8}$

When charges are placed in air, the electric force between them is:

$F_e=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}.......\left(1\right)$

However, when charge is placed in medium,

$F_e^{{}^{\prime }}=\frac{1}{4\pi ϵ}\frac{q_1{q}_2}{r^2}$

Where $ϵ={ϵ}_0\times {ϵ}_r$ is called permittivity of medium.

So, when charge placed in medium,

$4F=\frac{1}{4\pi {ϵ}_r{ϵ}_0}\frac{q_1{q}_2}{R^2}\dots \dots \left(2\right)$

From eqs.(1) and (2),

$4=\frac{r^2}{{ϵ}_r{R}^2}$ $(\because {ϵ}_r=16)$

$\Rightarrow R^2=\frac{r^2}{64}$

$\therefore R=\frac{r}{8}$

Hence, option $\left(c\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question ?}\\ \text{Bottom line: When charges are placed at same separation (r)}\\ \text{in a medium of dielectric constant}{ϵ}_r\text{the force between}\\ \text{them decreases.}\\ \text{i.e.}{F}^{{}^{\prime }}=\frac{F}{{ϵ}_r}\end{array}$