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

Explanation

Area of the plate of a parallel plate capacitor, `A=90cm^(2)=90xx10^(-4)m^(2)`
Distnace betweent he plates , d=2.5 mm =`2.5xx10^(-3) m`
potential difference across the plates , V=400 v
(a) capacitance of the capacitor is given by the relation.
`C=(in_(0)A)/d`
Electrostatic energy stored in the capacitor is given by the relation , `E_(1)=1/2 CV^(2)`
`=1/2 (in_(0)A)/d V^(2)`
where, `in_(0)`=permitivity of free space =`8.85xx10^(-12)C^(2)N^(-1)m^(-2)`
`:. E_(1)=(1xx8.85xx10^(-12)xx90xx(400)^(2))/(2xx2.5xx10^(-3))=2.55xx10^(-6)J`
Hence, the electrostatic energy stored by the capacitor is `2.55xx10^(-6)J`
(B) volume of the given capacitor
`V'=A xx d`
`=90xx10^(-4) xx25xx10^(-3)`
`=2.25xx10^(-4) m^(3)`
Energy stored in the capacitor per unit volume is given by
`u=(E_(1))/(V')`
`=(2.55xx10^(-6))/(2.25xx10^(-4)) =0.11 Jm^(-3)`
Again `u=(E_(1))/(v')`
`=(1/2 CV^(2))/(Ad) =((in_(0)A)/(2d)V^(2))/(Ad) =1/2 in_(0)(V/d)^(2)`
where,
`V/d`=electric intensity =E
`:. u=1/2 in_(0)E^(2)`