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)

$\frac{d}{\sqrt{k}}$

Answer 2

Electric force between two point charges $q_1$ and $q_2$ separated by a distance d in air is given by $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} \, \, \, \, \, \,$ ...(i) Electric force between two point charges $q_1$ and $q_2$ separated by a distance $d'$ in a medium of relative permittivity $k$ is given by $F' = \frac{1}{4\pi E_{0}k } . \frac{q_{1}q_{2}}{d'^{2}}$ According to question $, F = F'$ $\therefore \, \, \frac{1}{4\pi e_{0}} \frac{q_{1}q_{2}}{d^{2}} = \frac{ 1}{4\pi e_{0}k} \frac{q_{1}q_{2}}{d'^{2}}$ $\frac{1}{d^{2} } = \frac{1}{kd'^{2}}$ or $kd'^{2} = d^{2}$ or $d' = \frac{d}{\sqrt{k}} .$