Question

A parallel plate capacitor with plate area $A$ and separation distance between the plates $d$, is charged by a constant current $i$. Consider a plane surface of area $\frac{A}{3}$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

• A. $\frac{i}{3}$
• B. $\frac{i}{4}$
• C. $\frac{i}{2}$
• D. $i$

Answer 1

The correct option is A $\frac{i}{3}$

Let the charge at some instant be $Q$ on the capacitor.
Electric field between the plates of the capacitor is, $E=\frac{Q}{A{ϵ}_0}$
Flux passing through area $\frac{A}{3}$ will be
${\varphi }_E=E\left[\frac{A}{3}\right]=\frac{Q}{A{ϵ}_0}\left(\frac{A}{3}\right)=\frac{Q}{3{ϵ}_0}$
The displacement current,
$i_d={ϵ}_0\left[\frac{d{\varphi }_E}{dt}\right]=\frac{1}{3}\frac{dQ}{dt}=\frac{i}{3}$