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

To find the displacement current through the parallel plate capacitor, we will follow these steps:

### Step 1: Understand the given charge function
The charge \( Q \) on the capacitor is given by:
\[ Q = Q_0 \cos(2 \pi f t) \]
where \( Q_0 \) is the maximum charge, \( f \) is the frequency, and \( t \) is time.
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