A capacitor of capacitance `12.0 muF` is joined to an `AC` source of frequency `200 Hz`. The rms current in the circuit is `2.00 A`. (a) Find the rms voltage across the capacitor. (b) Find the average energy stored in the electric field between the plates of the capacitor.

Explanation

The impedance of the capacitor `= 1/(omegaC)`
`= 1/((2pi xx 200 s^(-1)) (12 muF)) = 66.3 Omega`.
The rms voltage across the capacitor
`= i_(rms) Z = 2.0 A xx 66.3 Omega = 133 V`.
(b) The energy stored in the electric field `= 1/2 CV^2`.
Hence the average energy stored `= 1/2 C vec V^2`.
But `vec V^2 = (V_(rms)^2`.
Thus, the average energy stored
`= 1/2 xx (12 muF) xx (133 V)^2 = 0.106 J`.