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The charge of a parallel plate capacitor is varying as $q= q_0 sin \, 2\pi ft$ . The plates are very large and close together (area= A, separation= d). Neglecting edge effects, the displacement current through the capacitor is |
$\frac{d}{Aε_0}$ $\frac{d}{ε_0}sin2 \pi ft$ $2\pi fq_0 cos 2 \pi ft$ $\frac{2\pi fq_0}{ε_0}cos2\pi ft$ |
$2\pi fq_0 cos 2 \pi ft$ |
$i = \frac{dq}{dt} =\frac{d}{dt} (q_0 sin 2 \pi ft) = q_02\pi f cos 2 \pi ft $ |
