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:

F/4

Explanation:

The electrostatic force between two charge q1 and q2 separated at distance d is

$F=\frac{1}{4(pi)e_{o} } \frac{q1q2}{d^{2} }$

If a medium of relativity in introduced then the electrostatic force is

$F'=\frac{1}{4(pi)ee_{o} } \frac{q1q2}{d^{2} }$ = F/e = F/4

Answer 2

Answer:

answer of this question is F/4