The plates of 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. How much electrostatic energy is stored by the capacitor?

Explanation

`A = 90 cm^(2) = 90 xx 10^(-4) m^(2)`
`d = 2.5 mm = 2.5 xx 10^(-3) m`
V = 400 V
`C = (epsilon_(0)A)/d`
`=(8.854 xx 10^(-12) xx 90 xx 10^(-4))/(2.5 xx 10^(-3))`
`= 318.74 xx 10^(-12-4+3)`
`C = 318.74 xx 10^(-13)`
`C = 3.19 xx 10^(-11) F`
Energy stored in the capacitor `= U = 1/2 CV^(2)`
`=1/2 xx 3.19 xx 10^(-11) xx (400)^(2)`
`=25.52 xx 10^(-7)`
`U = 2.55 xx 10^(-6) J`