A parallel plate air condenser whose each plate has an area `S = 100 cm^(2)` is connected in series to an ac circuit. Find the electric field strength amplitude in the capacitor if the sinusolidal current amplitude in lead wires is equal to `I_(m) = 1.0 mA` and the current frequency equals `omega = 1.6 . 10^(7)S^(-1)`.

Explanation

Given `I = I_(m) sin omega t`. We see that
`j = (I_(m))/(S) sin omega t = -j_(d) = -(del D)/(del t)`
or, `D = (I_(m))/(omega S) cos omega t`, so , `E_(m) = (I_(m))/(epsilon_(0) omega S)` is the amplitude of the electric field and is `7 V//cm`