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The charge on a parallel plate capacitor...

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.

A

`I_(d)=-2pivq_(0) sin2pivt`

B

`I_(d)=2pivq_(0) sin2pivt`

C

`I_(d)=2pivq_(0) cos2pivt`

D

`I_(d)=-2pivq_(0) cos2pivt`

Text Solution

Verified by Experts

The correct Answer is:
A
The displacement current through the capacitor is.
`I_(d)=I_(c)=(dq)/(dt)`
Here, `q=q_(0) cos2pivt` (given)
Putting this value in Eq (`i`), we get
`I_(d)=I_(c)=-q_(0) sin 2pivtxx2piv`
`I_(d)=I_(c)=-2pivq_(0)sin2pivt`
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