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 symmetrically between the plates. Find the displacement current through this area.

Answer 1

Given that,

Area $A=\frac{A}{2}$

Separation between the plates $=d$

Constant current $=i$

Now, the electric field between plates of the capacitor

$E=\frac{q}{{\epsilon }_{0}A}$

Now, the flux through the area

$\varphi E=\frac{q}{{\epsilon }_{0}A}\times \frac{A}{2}$

$\varphi E=\frac{q}{2{\epsilon }_0}$

Now, the displacement current is

$I_d={\epsilon }_0\times \frac{d\varphi E}{dt}$

$I_d={\epsilon }_0\times \frac{d\left(\frac{q}{2{\epsilon }_0}\right)}{dt}$

$I_d=\frac{1}{2}\left(\frac{dq}{dt}\right)$

$I_d=\frac{i}{2}$

Hence, the displacement current is $\frac{i}{2}$