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  • The charge on a parallel plate capacitor is varying as <img alt="\mathrm{q=q_0 \sin 2 \pi n t}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bq%3Dq_0%20%5Csin%202%20%5Cpi%20n%20t

The charge on a parallel plate capacitor is varying as \mathrm{q=q_0 \sin 2 \pi n t} The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:
 

Option: 1

\mathrm{\frac{q}{\varepsilon_0 \mathrm{~A}}}
 


Option: 2

\mathrm{\frac{q_0}{\varepsilon_0} \sin 2 \pi n t}
 


Option: 3

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

 


Option: 4

\mathrm{ \frac{2 \pi n \mathrm{q}_0}{\varepsilon_0} \cos 2 \pi \mathrm{nt} }


Answers (1)

best_answer

\mathrm{l}_{\mathrm{D}}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{q}_0 \sin 2 \pi \mathrm{nt}=2 \pi \mathrm{n} \mathrm{q}_0 \cos 2 \pi \mathrm{q}_0 \cos 2 \pi \mathrm{nt}

Hence option 3 is correct.

Posted by

Pankaj Sanodiya

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