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The charge on a parallel plate capacitor is varying as $q = q_0 sin 2\pi ft$ The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is: |
$ \frac{Q}{\epsilon_o A}$ $\frac{Q}{\epsilon_0} sin2\pi ft$ $2\pi fq_0 cos 2\pi ft$ $\frac{2\pi fq_0}{\epsilon_0} cos2\pi ft$ |
$2\pi fq_0 cos 2\pi ft$ |
$I_d = \frac{dq}{dt} = \frac{d}{dt} q_0 sin 2\pi ft = q_0 2\pi f cos 2\pi ft$ |
