Question

Two point charges exert on each other a force $F$ when they are placed $r$ distant apart in air. When they are placed $R$ distance apart in a medium of dielectric constant K, they were exerting the same force. The distance $R$ equals

• A. $\frac{r}{\sqrt{K}}$
• B. $\frac{r}{K}$
• C. $rK$
• D. $r\sqrt{K}$

Answer 1

The correct option is A $\frac{r}{\sqrt{K}}$

Given two point charges exert a force $F$ when placed $r$ apart in air. When they are placed $R$ apart inn a medium of dielectric constant $K$, the force between them is same. We have to find $R$.

For this, let $q_1$ and $q_2$ be two point charges. When these charges are placed in dielectric medium, then force is $\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{R^2}$

where $\frac{\epsilon }{{\epsilon }_0}=K=$ dielectric constant of the medium.

So, new force $=\frac{1}{4\pi {\epsilon }_{0}K}\frac{q_1{q}_2}{R^2}$

when placed in air, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

$\epsilon =$ permittivity of medium

${\epsilon }_0=$ permittivity of air

Given that force in both the cases is same; then $\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{r^2}=\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{K{R}^2}$

$r^2=K{R}^2$

So, $R=\frac{r}{\sqrt{K}}$