Ncert Exercises & Solutions

The charge on a parallel plate capacitor varies as $q = q_{0} \cos 2 πνt$ . The plates are very large and close together (area = A, separation =d). Neglecting the edge effects, find the displacement current through the capacitor.

Hint: The displacement current depends on the variation of charge.

Step 1: Find the rate of change of charge.

The displacement current through the capacitoris,

I_{d} = I_{c} = \frac{dq}{dt}

Here, q = q_{0} \cos 2 πνt (given)

Putting this vale in Eq (i), we get;

I_{d} = I_{c} = - q_{0} \sin 2 πνt \times 2 πν

I_{d} = I_{c} = 2 {πνq}_{0} \sin 2 πνt

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