Solve this:
Q. Two point charge $q_1 and q_2$ are placed at a distance of 50 cm from each other in air, and interact with a certain force. Now the same charges are put in an oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is :
(1) 16.6 cm.
(2) 22.3 cm.
(3) 35.0 cm.
(4) 28.4 cm.

Dear Student,
When two point charges are placed in air at distance 

$r$ then the Coulomb's force that acts in between them is given as,
 $F_{air}=\frac{1}{4\pi {\epsilon }_o}\frac{q_1{q}_2}{r^2}$
Now, if the same charges are placed in a dielectric having dielectric constant $k$ and separation of $d$, then the Coulomb's force in between them will be, 
 $F_{medium}=\frac{1}{4\pi {\epsilon }_{o}k}\frac{q_1{q}_2}{d^2}$
According to question, $F_{air}=F_{medium}$, therefore on equating both the equations we get,
 $$\begin{gathered} \frac{1}{4\pi {\epsilon }_o}\frac{q_1{q}_2}{r^2}=\frac{1}{4\pi {\epsilon }_{o}k}\frac{q_1{q}_2}{d^2} \\ \frac{1}{r^2}=\frac{1}{k}\frac{1}{d^2} \\ {r}^2=k{d}^2 \\ d=\sqrt{\frac{r^2}{k}} \end{gathered}$$
On substituting the values we get,
 $$\begin{gathered} d=\sqrt{\frac{{\left(50 cm\right)}^2}{5}} \\ d=\sqrt{\frac{2500 c{m}^2}{5}} \\ d=\sqrt{500 c{m}^2} \\ d=22.36 cm \end{gathered}$$
Thus, the two charges are placed at d= 22.36  cm in dielectric.
Option (b) is correct