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, $E=\frac{\sigma }{\epsilon _{0}}$ or, $\frac{q}{\epsilon _{0} A}$ .
Flux through the given area,
$\left(\phi\right)_{e}=\frac{q}{A \left(\epsilon \right)_{0}}\left(\frac{A}{4}\right)=\frac{q}{4 \left(\epsilon \right)_{0}}$ .
Displacement current,
$I_{D}=\epsilon _{0}\frac{d \phi_{E}}{d t}$
$=\left(\epsilon \right)_{0}\frac{d}{d t}\left(\frac{q}{4 \left(\epsilon \right)_{0}}\right)=\frac{i}{4}$ .