Q 2.5) A parallel plate capacitor with air between the plates has a capacitance of 8pF (1pF = 10-12 F. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

 

Given,

Capacitance, C = 8pF.

In first case the parallel plates are at a distance ‘d’  and is filled with air.

Air has dielectric constant, k = 1

Capacitance, C=k×ϵo×Ad=ϵo×Ad

Here,

A = area of each plate

 = permittivity of free space.

Now, if the distance between the parallel plates is reduced to half, then d1 = d/2

Given, dielectric constant of the substance, k1 = 6

Hence, the capacitance of the capacitor,

C1=k1×ϵo×Ad1=6ϵo×Ad/2=12ϵoAd

Taking ratios of eqns. (1) and (2), we get,

C1 = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.

Hence, capacitance between the plates is 96pF.