The plates of a parallel plate capacitor have an area of `100 cm^(2)` each and are separated by 3mm. The capacitor is charged by connecting it to a 400 V supply.
(a) Calculate the electrostatic energy stored in the capacitor.
(b) If a dielectric of dielectric constant 2.5 is introduced between the plates of the capacitor then find the electrostatic energy stored and also change in the energy stored.

Explanation

Given `A = 100 cm^(2) = 100 xx 10^(-4) m^(2), d = 3mm = 3 xx 10^(-3)m, V = 400 V, u_(1) = ?, K = 2.5, u_(2) = ?` Also `u_(2) - u_(1) = ?`
We have
`U = (1)/(2) C V^(2)`
`C = (epsilon_(0)A)/(d) = (8.854 xx 10^(-12) xx 100 xx 10^(-4))/(3 xx 10^(-3)) = (8.854 xx 10^(-16))/(3 xx 10^(-3)) = 295.13 xx 10^(-13)`
`C = 29.513 xx 10^(-12) F`
(i) `U_(1) = (1)/(2)CV^(2)`
`= (1)/(2) 29.513 xx 10^(-12) xx (400)^(2) = (4722080)/(2) xx 10^(-12) = 2361040 xx 10^(-12)`
`U_(1) = 23.610 xx 10^(-7) J`
(ii) If dielectric medium K = 2.5 is introduced between the plates of the capacitors, the its capacitance is
`C = (epsilon_(0) KA)/(d) = (8.854 xx 10^(-12) xx 2.5 xx 100 xx 10^(-4))/(3 xx 10^(-3)) = (2213.5)/(3 xx 10^(-3)) xx 10^(16) = 737.833 xx 10^(-13)`
`C = 73.7833 xx 10^(-12) F`
`U_(2) = (1)/(2) C V^(2) = (1)/(2) 73.7833 xx 10^(-12) xx (400)^(2) = 5902661 xx 10^(-12)`
`U_(2) = 59.02664 xx 10^(-7)J`
Change in energy
`Delta U = U_(2) - U_(1) = 59.02664 xx 10^(-7) - 23.610 xx 10^(-7)`
`Delta U = 35.14661 xx 10^(-7)J`