Question

Two equal charges separated by a distance R in air repel with a force F. At what distancethese charges are to be placed in an oil of relative permittivity 5 so that they repel withthe same force ?​

Answer 1

Answer:

R√(1/5).

Explanation:

Since, for the two particles will undergo electrostatic forces of attraction and  also we know that the permittivity of space is 1 but for oil it is given in the question is 5. So, we get that ε1/ε2 will be 1/5. We know that the force of  the electrostatic attraction is given by F= q1q2/4πεR^2 so, from here we get that the radius R² is inversely proportional to the 1/ε.

So, we get that the R²/R2² = ε2/ε1. Hence, this equation can be written as the distance where it must be placed or R = R2√ε1/ε2 which will be on solving as R2 = R√(1/5).