The plates of a parallel plate capacitor have an area of 90 cm^{2} each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

**(a)** How much electrostatic energy is stored by the capacitor?

**(b)** View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume *u*. Hence arrive at a relation between *u *and the magnitude of electric field *E *between the plates.

#### Solution

Area of the plates of a parallel plate capacitor, *A* = 90 cm^{2 }= 90 × 10^{−4 }m^{2}

Distance between the plates, *d* = 2.5 mm = 2.5 × 10^{−3 }m

Potential difference across the plates, *V* = 400 V

**(a) **Capacitance of the capacitor is given by the relation,

`C=(in_0A)/d`

Electrostatic energy stored in the capacitor is given by the relation, `E_1=1/2CV^2`

`=1/2 (in_0A)/d`

Where,

`in_0` = Permittivity of free space = 8.85 × 10^{−12} C^{2} N^{−1} m^{−2}

`E=(1.8.85xx10^-12xx90xx10^-4xx(400)^2)/(2xx2.5xx10^-3)=2.55xx10^-6 J`

Hence, the electrostatic energy stored by the capacitor is `2.55 xx 10^-6 J`

**(b) **Volume of the given capacitor,

`V'=Axxd`

`=90xx10^-4xx25xx10^-3`

`=2.25 xx10^-4 m^3`

Energy stored in the capacitor per unit volume is given by,

`u=E_1/(V')`

`=(2.55xx10^-6)/(2.25xx10^-4)=0.113 Jm^-3`

Again, u=`E_1/(V')`

`=((1/2)CV^2)/(Ad)=((in_0A)/(2d)V^2)/(Ad)=1/2in_0(V/d)^2`

Where,

`V/d`= Electric intensity = *E*

`therefore u=1/2 in_0E^2`