Question

Two point charges in air, at a distance of $20\text{cm}$ from each other, interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity $5$ to obtain the same magnitude of force of interaction?

• A. $8.94\times {10}^{-2}\text{m}$
• B. $7.94\times {10}^{-2}\text{m}$
• C. $6.94\times {10}^{-2}\text{m}$
• D. $5.94\times {10}^{-2}\text{m}$

Answer 1

The correct option is A $8.94\times {10}^{-2}\text{m}$

We know, electrostatic force,
$F=\frac{1}{4\pi {ϵ}_r{ϵ}_o}\frac{q_1{q}_2}{r^2}$

For air, ${ϵ}_r=1$
For given oil, ${ϵ}_r=5$

As magnitude of force of interaction is the same in both the medium,
$\frac{1}{r^2}=\frac{1}{{ϵ}_r{r}_o^{{}^2}}$
$[$On equating both the forces$]$

$\Rightarrow \frac{1}{{20}^2}=\frac{1}{5\times r_o^{{}^2}}$
$\Rightarrow r_o=8.94\text{cm}=8.94\times {10}^{-2}\text{m}$