Question

Two point charges $q_1$ and $q_2$ are placed at a distance of $50\text{m}$ from each other in air, interact with a certain electrostatic force. If the medium is replaced with oil whose relative permittivity is $5$, then the interaction force between them is still the same. Now the separation between them is

• A. $16.6\text{m}$
• B. $22.3\text{m}$
• C. $28.4\text{m}$
• D. $25.0\text{m}$

Answer 1

The correct option is B $22.3\text{m}$

Given that,
Initial distance between the charges, $r_0=50\text{m}$
Relative permittivity of oil, ${ϵ}_r=5$
When two charges are placed in air, the electric force between them is
$F_0=\frac{k{q}_1{q}_2}{r_0^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_0^2}$

Let the permittivity of the medium is ${ϵ}_m$
$\Rightarrow {ϵ}_m={ϵ}_0{ϵ}_r$
$\Rightarrow {ϵ}_m=5{ϵ}_0$
The electrostatic force in oil is given by
$\Rightarrow F=\frac{1}{4\pi {ϵ}_m}\frac{q_1{q}_2}{r^2}$

To maintain the same interacting force,
$F_0=F$
$\Rightarrow \frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_0^2}=\frac{1}{4\pi {ϵ}_r{ϵ}_0}\frac{q_1{q}_2}{r^2}$
or, $r^2=\frac{r_0^2}{{ϵ}_r}$
$\Rightarrow r=\frac{r_0}{\sqrt{{ϵ}_r}}$
Substituting values from question,
$\Rightarrow r=\frac{50}{\sqrt{5}}$
$\therefore r\approx 22.36\text{m}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Whenever Coulomb's law is applied in medium, we take}\\ {ϵ}_m={ϵ}_0{ϵ}_r\text{for the medium.}\\ \text{In medium electrostatic force between charges decreases,}\\ \text{thus to maintain same force, the separation between them must be decreased.}\end{array}$