The plates of a paralllel 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

(a) Electrostatic energy stored in the capacitor can be calculated as:
`U=(1)/(2)CV^(2)=(1)/(2) (epsilon_(0)A)/(d) V^(2)`
`=(1)/(2) ((8.85xx10^(-12))xx(90xx10^(-4))xx(400)^(2))/(2.5xx10^(-3))`
`=2.55xx10^(-6)J`
(b) Energy per unit volume is given as
`u=(U)/(Ad)=(2.55xx10^(-6)J)/(225xx10^(-7)m^(3))=0.113J m^(-3)`
Also, `u=("Energy")/("Volume")=(1)/(2)(epsilon_(0)A*V^(2))/(d xx A xx d)`
`=(1)/(2) (epsilon_(0)*V^(2))/(d^(2))=(1)/(2)epsilon_(0)*E^(2)`.