(B) The displacement current $I_D$ in a capacitor is equal to the rate of change of charge on the plates,which is the conduction current $I_C$.Given t

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$-q_0 \, 2\pi \nu \sin(2\pi \nu t)$

(B) The displacement current $I_D$ in a capacitor is equal to the rate of change of charge on the plates,which is the conduction current $I_C$.Given the charge $q = q_0 \cos(2\pi \nu t)$.The displacement current is given by $I_D = \frac{dq}{dt}$.$I_D = \frac{d}{dt} [q_0 \cos(2\pi \nu t)]$.Using the chain rule for differentiation,$\frac{d}{dt} \cos(kt) = -k \sin(kt)$.$I_D = q_0 \cdot (-2\pi \nu) \sin(2\pi \nu t)$.$I_D = -q_0 \, 2\pi \nu \sin(2\pi \nu t)$.

Class 12 Physics Electromagnetic Waves Maxwell's equations , Concept of displacement current and Hertz experiment

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The charge on a parallel plate capacitor varies as $q = q_0 \cos(2\pi \nu t).$ The plates are very large and close together (area $= A,$ separation $= d$). The displacement current through the capacitor is

A
$q_0 \, 2\pi \nu \sin(2\pi \nu t)$
B
$-q_0 \, 2\pi \nu \sin(2\pi \nu t)$
C
$q_0 \, 2\pi \sin(\pi \nu t)$
D
$q_0 \, \pi \nu \sin(2\pi \nu t)$

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Electromagnetic Waves

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Maxwell's equations , Concept of displacement current and Hertz experiment

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The charge on a parallel plate capacitor varies as $q = q_0 \cos(2\pi \nu t).$ The plates are very large and close together (area $= A,$ separation $= d$). The displacement current through the capacitor is

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