# An air-filled parallel plate c

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

a. Calculate the dielectric constant of the insertedmaterial.

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

Answer:

Given,

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

The capacitance of the dielectric inserted parallel platecapacitor ( ) = 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 thecapacitor ( V ) = 6.0 V

The charge on the capacitor is given as,

As the modifications were made on the capacitor after removingthe battery. So, the capacitor will have the same charge which isbeing 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 platecapacitor ( ) =

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

putting the values,

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