Answer:
The force is now 0.5 N.
Explanation:
Force = 1.5 N
dielectric constant , k = 3
Let the two charges are q and q' and the distance between them is r.
The electrostatic force between the two charges is given by
[tex]F \alpha \frac{ q q'}{r^2}..... (1)[/tex]
When a dielectric material is inserted between the two charges, the new force is
[tex]F' \alpha \frac{ q q'}{kr^2}..... (2)[/tex]
From (1) and (2)
F' = F/K = 1.5/3 = 0.5 N