A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of the same area but the thickness, d 2 , is inserted between the plates as shown in figure having dielectric constant K ( = 4 ) . The ratio of new capacitance to its original capacitance will be,

Step-by-step solution

First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates, C 0 = ϵ 0 A d , when a dielectric slabs of thickness, t and dielectric constant k , inserted between plates then-new capacitance becomes, C k = ϵ 0 A d - t + t k ⇒ C k = ϵ 0 A d - d 2 + d 8 , here t = d 2 . ⇒ C k = 8 5 ϵ 0 A d ⇒ C k = 8 5 C 0 ⇒ C k C 0 = 8 5