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

I/2

Charge on capacitor plates at time t is, q = It.

Electric field between the plates at this instant is

$\mathrm{E}=\frac{\mathrm{q}}{\mathrm{A} \varepsilon_0}=\frac{\mathrm{It}}{\mathrm{A} \varepsilon_0}$                  (i)

Electric flux through the given area A/2 is

$\phi_{\mathrm{E}}=\left(\frac{\mathrm{A}}{2}\right) \mathrm{E}=\frac{\mathrm{It}}{2 \varepsilon_0}$    (Using (i))        (ii)

Therefore, displacement current

$\mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{It}}{2 \varepsilon_0}\right)=\frac{\mathrm{I}}{2}$       (Using (ii))