Question

Two point charges $q_1$ and $q_2$ are placed at a distance of $50\text{m}$ from each other in air, and interact with a certain force $F$. The same charges are now put in oil whose relative permittivity is $5.0$. Find the new force of interaction between them :

• A. $F$
• B. $\frac{F}{5}$
• C. $5F$
• D. $\frac{F}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{F}{5}$

Given,
Relative permittivity of oil $\left(k\right)=5.0$

Since the electrostatic force in both the situations is the same, we can take $r_1$ as the initial separation and $r_2$ as the separation when the charges are put in oil.

If the electrostatic force between two charges in vacuum is $F$, then the electrostatic force between the same two charges in a medium of relative permittivity $k$ is $\frac{F}{k}$
​​​​​​​
So, new force of interaction between the charges $=\frac{F}{5}$

Hence, option (b) is the correct answer.