Question

A slab of material of dielectric constant
k has the same area A as the plates of a
parallel plate capacitor and has thickness
(3/4d), where d is the separation of the
plates. The change in capacitance when
the slab is inserted between the plates is​

Answer 1

Explanation:C= dAϵ 0 C ′ = d−t+ kt Aϵ 0 Put t= 43d ;C ′ = 3+k4k . dAϵ 0 C ′ = 3+k4k .C

am i right,