#### Question

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?

#### Solution

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was *d *and it was filled with air. Dielectric constant of air, *k*= 1

Capacitance, *C*, is given by the formula,

`C=(kin_0A)/d`

`=(in_0A)/d` ...(1)

Where,

*A* = Area of each plate

`in_0`= Permittivity of free space

If distance between the plates is reduced to half, then new distance, *d*^{’} = `d/2`

Dielectric constant of the substance filled in between the plates, k' = 6

Hence, capacitance of the capacitor becomes

`C'=(k'in_0A)/d=(6in_0A)/(d/2)` ...(2)

Taking ratios of equations (i) and (ii), we obtain

`C'=2xx6C`

=12C

=12 x 8=96 pF

Therefore, the capacitance between the plates is 96 pF.