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The charge on a parallel plate capacitor varies as q = q0 cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects

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प्रश्न

The charge on a parallel plate capacitor varies as q = q0 cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?

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उत्तर

This displacement current through the capacitor is given by,

`I_d = I_c = (dq)/(dt)`  ......(i)

Here we are given, q = q0 cos 2πνt

Putting this value in equation (i), we get

`I_d = I_c = - q_0 sin 2pivt xx 2 piv`

`I_d = I_c = - 2 pivq_0  sin  2pivt`

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पाठ 8: Electromagnetic Waves - MCQ I [पृष्ठ ५०]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 8 Electromagnetic Waves
MCQ I | Q 8.16 | पृष्ठ ५०

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