Question

A parallel plate capacitor with plate area $A$ and separation between plates $d$ , is charged by a constant current $i$ . Consider a plane surface of area $\frac{A}{4}$ parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is,

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

Answer 1

Answer: (C)

The electric field between plates of the capacitor is given by, $E=\frac{\sigma }{{ϵ}_0}$ or, $\frac{q}{{ϵ}_{0}A}$ .
Flux through the given area,
${\left(\varphi \right)}_e=\frac{q}{A{\left(ϵ\right)}_0}\left(\frac{A}{4}\right)=\frac{q}{4{\left(ϵ\right)}_0}$ .
Displacement current,
$I_D={ϵ}_0\frac{d{\varphi }_E}{dt}$
$={\left(ϵ\right)}_0\frac{d}{dt}\left(\frac{q}{4{\left(ϵ\right)}_0}\right)=\frac{i}{4}$ .