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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.

Text Solution

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Displacement current, `I_D`=conduction current `I_C`.
`:. I_D=I_C=(dq)/(dt)=d/(dt)[q_0 cos 2pivt]=-q2piv sin 2pivt.`
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