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The charge on a parallel plate capacitor is varying as $q = q_{0} \sin (2 π n t)$ . The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:

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$\frac{q}{ε_{0} A}$

$2 π n q_{0} c o s (2 π n t)$

$\frac{q_{0}}{ε_{0}} \sin (2 π n t)$

$\frac{2 π n q_{0}}{ε_{0}} c o s (2 π n t)$

answer is C.

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Detailed Solution

$I_{D} = \frac{d q}{d t} = \frac{d}{d t} q_{0} \sin (2 π n t) = 2 π n q_{0} \cos (2 π n t)$

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