The plates of a parallel plate capacitor have an area of `90 cm^(2)` each and are separated by 2.5mm. The capacitor 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 hte 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

`90 cm^(2)= 90 xx 10^(-4) m^(2)`
`d=2.5mm =2.5 xx 10^(-3)m`
V=400V
a. Energy, `E=1/2CV^(2)=1/2 xx (8.85 xx 10^(-12) xx 90 xx 10^(-4))/(2.5 xx 10^(-3)) xx 400 xx 400`
`=(8.85 xx 10^(-8) xx 9 xx 8)/(2.5)=(88.5 xx 72 xx 10^(-8) xx 4)/(100)=2.55 xx 10^(-6)J`
b. `u=(1/2 CV^(2))/(A xx d)=1/2 xx (epsi_(0) A xx V^(2))/(d xx Ad)=1/2epsi_(0) E^(2)`