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\varepsilon_{0}} = \frac{It}{A\varepsilon _{0}}\quad ...\left(i\right)$
Electric flux through the given area $\frac{A}{2}$ is
$\phi_{E} = \left(\frac{A}{2}\right) E = \frac{It}{2\varepsilon _{0}} \quad$ Using $\left(i\right) \quad ...\left(ii\right)$
Therefore, displacement current
$I_{D} = \varepsilon_{0} \frac{d\phi_{E}}{dt}$
$= \varepsilon _{0} \frac{d}{dt}\left(\frac{It}{2\varepsilon _{0}}\right) = \frac{I}{2}\quad$ [Using $\left(ii\right)$]