A slab of material of dielectric constant k has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, when d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Explanation

Before insertion of the dielectric slab,
`C_1= (in_0 alpha)/(d) " "[alpha = " area of each plate "]`
After insertion of a dielectric slab of thickness t,
`C_2 = (in_0 alpha)/(d-t(1-(1)/(k))) = (in_0 alpha)/(d-(d)/(2)(1-(1)/(k)))[ :.t= (d)/(2)]`
`= (in_0 alpha)/(d((1)/(2)+(1)/(k)))= (C_1)/((1)/(2)+(1)/(k))[:. C_1= (in_0 alpha)/(d)]`.