Question

Two point charges in air ,at a distanceof 20 cm from each other interact with a certain force. At what distance from each other should these charges be placed in oil having a dielectric constant 5, to obtain the same force of interaction?

Answer 1

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,
$$\begin{gathered} d=\sqrt{\frac{{\left(20cm\right)}^2}{5}} \\ d=\sqrt{\frac{400c{m}^2}{5}} \\ d=\sqrt{80c{m}^2} \\ d=8.94cm \\ d\approx 9cm \end{gathered}$$
Thus, the two charges are placed at $d\approx 9cm$ in dielectric.