Question

A capacitor made of two parallel places each of plate area A and separations d, being charged by an external ac source. How the displacement current inside the capacitor is same as the current charging the capacitor

Answer 1

TD=ε0dμEdtlet the altering emf charging the plates of capacitor be $E=E_{0}sin\omega t$

charge on the capacitor $q=EC={CE}_{0}sin\omega t$

$I=\frac{dq}{dt}=\frac{d}{dt}\left({CE}_{0}sin\omega t\right)=\omega {CE}_{0}cos\omega t=I_{0}cos\omega t(where{I}_0=\omega {CE}_0)$

Displacement current

$I_D={\epsilon }_0\frac{d{\psi }_E}{dt}={\epsilon }_{0}A\frac{dE}{dt}={\epsilon }_{0}A\frac{d\left(\frac{q}{{\epsilon }_{0}A}\right)}{dt}={\epsilon }_{0}A\frac{d}{dt}\left(\frac{{CE}_{0}sin\omega t}{{\epsilon }_{0}A}\right)$

$=\frac{d}{dt}\left({CE}_{0}sin\omega t\right)=\omega {CE}_{0}cos\omega t=I_{0}cos\omega t$

Thus the displacement current inside the capacitor is the same as current charging the capacitor.