question_answer

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

Answer:

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. \[I=\frac{dq}{dt}=\frac{d}{dt}(C{{E}_{0}}sin\omega t)=\omega C{{E}_{0}}\cos \omega t={{I}_{0}}\cos \omega t\]                                     \[[where,\,\,{{I}_{0}}=\omega C{{E}_{0}}]\] Displacement current, \[{{I}_{d}}={{\varepsilon }_{0}}\frac{d{{\phi }_{E}}}{dt}\] \[{{\varepsilon }_{0}}A\frac{d(E)}{dt}={{\varepsilon }_{0}}A\frac{d}{dt}\left( \frac{q}{{{\varepsilon }_{0}}A} \right)={{\varepsilon }_{0}}A\frac{d}{dt}\left( \frac{C{{E}_{0}}\sin \omega t}{{{\varepsilon }_{0}}A} \right)\] \[=\frac{d}{dt}(C{{E}_{0}}sin\omega t)=\omega C{{E}_{0}}\cos \omega t={{I}_{0}}\cos \omega t\] Thus, the displacement current inside the capacitor is the same as the current charging the capacitor.