The charge on a parallel plate capacitor varies as $q = q_{0} \cos (2 π v t)$ .The plates are very large and close together (Area = $A$ , separation = $d$ ). The displacement current through the capacitor is:

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$2 q_{0} π \sin (π v t)$

$- 2 q_{0} π v \sin (2 π v t)$

$2 q_{0} π v \sin (π v t)$

$2 q_{0} π v \sin (2 π v t)$

answer is B.

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Detailed Solution

Displacement current = conduction current

$∴ i_{d} = \frac{d q}{d t} = \frac{d}{d t} [q_{0} \cos (2 π v t)] = - q_{0} 2 π v \sin (2 π v t)$

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