The charge on a parallel plate capacitor vari | vedclass

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Question

(B) The displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ flowing through the connecting wires,which is defined as th

Vedclass · Accepted Answer

$I_d = -2\pi 
u q_0 \sin(2\pi 
u t)$

Vedclass · Answer

(B) The displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ flowing through the connecting wires,which is defined as the rate of change of charge on the plates.$I_d = I_c = \frac{dq}{dt}$Given the charge $q = q_0 \cos(2\pi 
u t)$,we differentiate with respect to time $t$:$I_d = \frac{d}{dt} [q_0 \cos(2\pi 
u t)]$Using the chain rule,the derivative of $\cos(kx)$ is $-k \sin(kx)$:$I_d = q_0 \cdot [-\sin(2\pi 
u t)] \cdot (2\pi 
u)$$I_d = -2\pi 
u q_0 \sin(2\pi 
u t)$

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 $= d$). Neglecting the edge effects,find the displacement current through the capacitor.

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If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$,then the area of one plate of the capacitor is ($\varepsilon_{0}$ is permittivity of free space).

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