Question

A parallel plate capacitor with plate area $A$ and separation between the plates $d$, is charged by a constant current $I$ .Consider a plane surface of area $A/2$ parallel to the plates and drawn between the plates. The displacement current through the area is

• A. $I$
• B. $\frac{I}{2}$
• C. $\frac{I}{4}$
• D. $\frac{I}{8}$

Answer 1

Answer: (B)

Charge on capacitor plates at time $t$ is, $q=$ It. Electric field between the plates at this instant is
$E=\frac{q}{A{\epsilon }_0}=\frac{It}{A{\epsilon }_0}...\left(i\right)$
Electric flux through the given area $\frac{A}{2}$ is
${\varphi }_E=\left(\frac{A}{2}\right)E=\frac{It}{2{\epsilon }_0}$ Using $\left(i\right)...\left(ii\right)$
Therefore, displacement current
$I_D={\epsilon }_0\frac{d{\varphi }_E}{dt}$
$={\epsilon }_0\frac{d}{dt}\left(\frac{It}{2{\epsilon }_0}\right)=\frac{I}{2}$ [Using $\left(ii\right)$]