The plates or 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, und obtain the energy per unit volume "· Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Explanation

(a) Here A = 90 `cm^(2) = 90xx10^(-4) m^(2) = 9xx10^(-3) m^(2)`
d = 2.5 mm = `2.5 xx10^(-3)` m
`in_(0) = 8.85 xx10^(-12) Fm^(-1)`
V = 400 V
Capacitance of parallel of parallel of parallel plate capacitor
`C= (in_(0)A)/(d)=(8.85xx10^(-12)xx9xx10^(-3))/(2.5xx10^(-3))`
`:. C = 31.86xx10^(-12) F = 31.86 pF`
Energy stored in capacitor
`U = (1)/(2) CV^(2) =(1)/(2) xx31.86 xx10^(-12)xx(400)^(2)`
`:. U = 254.88xx10^(-8) `J
`:. U = 2.55 xx10^(-6) J = 2.55 mu J`
(b) Enerby per unit of volume or energy density in capacitor
`rho_(E) = u = (U)/(Ad) = (2.55xx10^(-6))/(9xx10^(-3)xx2.5xx10^(-3))`
`:. u =0.113 jm^(-3)`
Relation between `rho_(E)(u)` and E
`rho_(E) (U)/(Ad) =(1//2CV^(2))/(Ad)`
`rho_(E)=(1)/(2)(in_(0)A)/(d) xx(V^(2))/(Ad)[because C = (in_(0)A)/(d)]`
`:. rho_(E) = (1)/(2) in_(0)((V^(2))/(d^(2)))`
`:. rho_(E)=(1)/(2)in_(0)E^(2)[because(V)/(d)=E]`