Question

A capacitor, made of two parallel plates each of plate area $A$ and separation $d$, being charged by an external AC source. Show that the displacement current inside the capacitor is same as the current charging the capacitor.

Answer 1

Let the alternating emf charging the plates of capacitor be $E=E_{0}sin\omega t$.
Charge on the capacitor.
$q=EC=C{E}_{0}sin\omega t$
Instantaneous current is $I$.
$I=\frac{dq}{dt}=\frac{d}{dt}\left(C{E}_{0}sin\omega t\right)$
$=\omega C{E}_{0}cos\omega t$
$=I_{0}cos\omega t$ ... (i)
where, $I_0=\omega C{E}_0$
Displacement current,
$I_D={\epsilon }_0\frac{d{\varphi }_E}{dt}$
$={\epsilon }_{0}A\frac{d\left(E\right)}{dt}[\because \varphi =E.A]$
$={\epsilon }_{0}A\frac{d}{dt}\left(\frac{q}{{\epsilon }_{0}A}\right)[\because E=\frac{q}{{\epsilon }_{0}A}]$
$={\epsilon }_{0}A\frac{d}{dt}\left(\frac{C{E}_{0}sin\omega t}{{\epsilon }_{0}A}\right)$
$=\frac{d}{dt}\left(C{E}_{0}sin\omega t\right)$
$=\omega C{E}_{0}cos\omega t$
$=I_{0}cos\omega t...\left(ii\right)$
From equations (i) and (ii), the displacement current inside the capacitor is same as the current charging the capacitor.