The charge on a parallel plate capacitor varies as `=q_0 cos 2pi ft`. The plates are very large and close together (area=a,separation=d). Neglecting the edge effects, find the displacement current through the capacitor.

Explanation

the displacement current through the capacitor is ,
`I_(d) =I_(c) =(dq)/(dt)`
`"Here"" " q=q_(0) " cos " 2pivt ("given")`
Putting this value in Eq (i), we get
`I_(d) =I_(c) =- q_(0) " sin"2 pi vt xx 2 piv`
`I_(d) =I_(c) =- 2pivq_(0) " sin"2pivt`