Question

The electric force between two point charges separated by a certain distance d in air is F.The distance at which they should be placed in a medium of relative permittivity k so that the force remain the same is

• A. $d$
• B. $\frac{d}{k}$
• C. $kd$
• D. $\frac{d}{\sqrt{k}}$

Answer 1

Answer: (D)

Electric force between two point charges $q_1$ and $q_2$ separated by a distance d in air is given by
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^2}$ ...(i)
Electric force between two point charges $q_1$ and $q_2$ separated by a distance $d^{\prime }$ in a medium of relative permittivity $k$ is given by
$F^{\prime }=\frac{1}{4\pi E_{0}k}.\frac{q_1{q}_2}{d^{\prime 2}}$
According to question $,F=F^{\prime }$
$\therefore \frac{1}{4\pi e_0}\frac{q_1{q}_2}{d^2}=\frac{1}{4\pi e_{0}k}\frac{q_1{q}_2}{d^{\prime 2}}$
$\frac{1}{d^2}=\frac{1}{k{d}^{\prime 2}}$ or $k{d}^{\prime 2}=d^2$
or $d^{\prime }=\frac{d}{\sqrt{k}}.$