A parallel plate capacitor with air between the plates has a capacitance of 8 pF (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: The capacitance of a parallel plate capacitor is 8 pF.

Capacitance of the capacitor is given as,

C= k ε 0 A d

where, capacitance of the capacitor C, area of each plate is A, permittivity of free space is ε 0 distance between the parallel plates is d and dielectric constant is k.

By substituting the given values, we get

C= ε 0 A d

If the distance between the plates is reduced by half and space between them is filled with dielectric constant 6.

The capacitance of the capacitor is given as,

C'= 6 ε 0 A d/2 =12×C

By substituting the given values in the above equation, we get

C'=12×C =12×8 =96 pF

Thus, if the distance between the plates is reduced by half and space between them is filled with dielectric constant 6 then capacitance of the capacitor is 96 pF.