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The charge of a parallel plate capacitor is varying as \mathrm{ q=q_0 \sin 2 \pi f t }. The plates are very large and close together ( area =A, separation =d). Neglecting edge effects, the displacement current through the capacitor is

Option: 1

\mathrm{\frac{d}{A \varepsilon_0}}


Option: 2

\mathrm{\frac{d}{\varepsilon_0} \sin 2 \pi f t }


Option: 3

\mathrm{2 \pi f q_0 \cos 2 \pi f t }


Option: 4

\mathrm{\frac{2 \pi f q_0}{\varepsilon_0} \cos 2 \pi f t }


Answers (1)

best_answer

\mathrm{i=\frac{d q}{d t}=\frac{d}{d t}\left(q_0 \sin 2 \pi f t\right)=q_0 2 \pi f \cos 2 \pi f t }

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Rishabh

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