A slab of dielectric constant K has the same crosssectional area as the plates of a parallel plate capacitor and thickness 3 4 d , where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given C 0 = capacitance of capacitor with air as medium between plates.)

Step-by-step solution

From the above diagram, we can write, x + y + 3 d 4 = d x + y = d 4 Initially, when there is no dielectric inserted between the plates the capacitance will be, C 0 = ϵ 0 A   d . When the dielectric slab is inserted, we can assume there are 3 capacitors that are connected in series. C 1 = ϵ 0 A x , C 2 = 4 K ϵ 0 A 3 d and C 3 = ϵ 0 A y The equivalent capacitance, 1 C = 1 C 1 + 1 C 2 + 1 C 3 ⇒ 1 C = 1 ϵ 0 A x + 3 d 4 K + y ⇒ 1 C = 1 ϵ 0 A d 4 + 3 d 4 K ⇒ C =