Question

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

• A. $q_{0}2\pi\upsilon\sin\pi\upsilon t$
• B. $- q_{0}2\pi\upsilon\sin 2\pi\upsilon t$
• C. $q_{0}2\pi\sin\pi\upsilon t$
• D. $q_{0}\pi\upsilon\sin 2\pi\upsilon t$

Answer 1

Answer: (B)

$- q_{0}2\pi\upsilon\sin 2\pi\upsilon t$

Answer 2

: Displacement current, $I_{D} =$conduction current, $I_{C}$

$\therefore\frac{dq}{dt} = \frac{d}{dt}\lbrack q_{0}\cos 2\pi\upsilon t\rbrack$

$= - q_{0}2\pi\upsilon\sin 2\pi\upsilon t$