Question

.Two point charges placed at a distances of 20 cm in air repel each other with a certain force. When a dielectric slab of thickness 8 cm and dielectric constant K is introduced between these point charges, force of interaction becomes half of it’s previous value. Then K is approximately

Answer 1

let the two charges be q1,q2 and the distance between them be r = 0.2m(20 cm)
by coulomb's law F = K (q1q2/r^2)
when the dielectric is introduced f(force in medium) is

                             f = F/2 = [K/ε(q1q2/r^2)]/2 (ε→dielectric constant)

∴F/f =F/(F/2)=2 = 2ε
            ⇒ε=1