The plates of a paralllel plate capacitor have an area of `90 cm^(2)` each and are separated by 2.5mm. The capacitane 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 realtion between U and the magnitude of electric field E between the plates.

Explanation

Here, `A = 90 cm^(2) = 90xx10^(-4) m^(2) = 9xx10^(-3) m^(2), d = 2.5mm = 2.5xx10^(-3)m, V = 400 volt,U = ?`
`U = (1)/(2) CV^(2) = (1)/(2) (in_(0) A)/(d) V^(2) , U = (8.85xx10^(-12)xx9xx10^(-3)(400)^(2))/(2xx2.5xx10^(-3)) = 2.55xx10^(-6) J`
(b) Volume of capacitor, `V = A xx d = 90xx10^(-4)xx25xx10^(-3) m^(3) = 2.25x10^(-4) m^(3)`
`En ergy//volume = U = (2.55xx10^(-6))/(2.25xx10^(-4)) = 0.113 J//m^(3) As U = (u)/(V) = ((1)/(2) CV^(2))/(Ad) = ((in_(0) A)/(2d) V^(2))/(Ad) = (1)/(2) in_(0) ((V)/(d))^(2)`
But `(V)/(d) = E`, electric intensity `:. U = (1)/(2) in_(0) E^(2)`