Question

Two charges q1 and q2 are placed in vacuum at a distance d and force between
them is F. If a medium of relative permittivity 4 is introduced between them, then
the new force will be
A. F/4
B. F/2
C. 2F
D. 4F

Answer 1

Answer:

Force between the charges become $\frac{F}{4}$

So option (a) is correct answer

Explanation:

We have given two charges $q_1\ and\ q_2$ which are separated by a distance d and force between them is F

Now we are introducing a medium of relative permittivity of 4

We know that force between to charges is inversely proportional to relative permittivity

So force between the charges become $\frac{F}{4}$

So option (a) is correct answer