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:
Answer & explanation
Correct answer: option 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))