Question

An air-filled parallel plate capacitor has a capacitance of 4.40 μF. The plate spacing is now doubled and a dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 16.2 μF.

a. Calculate the dielectric constant of the inserted material.

b. If the original capacitor was charged to a potential difference of 6.0 V and the battery was disconnected when the modifications to the capacitor was made, by what factor did the energy stored in the capacitor change due to this modification?

Answer 1

Given,

The capacitance of the air-filled parallel plate capacitor ( ) = 4.40 μF

The capacitance of the dielectric inserted parallel plate capacitor ( ) = 16.2 μF

Capacitance of any parallel plate capacitor is given as,

where, K is the dielectric constant of the material

is the permittivity of the air

A is the area the parallel plate

d is the distance between the parallel plates.

a. We know that the dielectric constant of air = 1

Assume the dielectric constant of the material = k'  

Therefore,   and   

So, we have   ,

putting the values of capacitances,

  

The dielectric constant of the inserted material = 3.68.

b. The potential difference across the capacitor to charge the capacitor ( V ) = 6.0 V

The charge on the capacitor is given as,  

As the modifications were made on the capacitor after removing the battery. So, the capacitor will have the same charge which is being after charging to a potential difference of 6.0 V.

Charge on the capacitor,

putting the values,

Energy of any parallel plate capacitor is given as,

where, Q is the electric charge of the capacitor

C is the capacitance of the capacitor

The energy of the air-filled parallel plate capacitor ( ) =

The capacitance of the dielectric inserted parallel plate capacitor ( ) =

The factor by which the energy stored in the capacitor change due to this modification,

  

putting the values,

So, the factor of the energy stored after modification = 0.27