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, the displacement current through this area, will be

Question

Dr. Abubacker · Accepted Answer

(b) Suppose the charge on the capacitor at time $t$ is $Q$, the electric field between the plates of the capacitor is $E = \frac{Q}{\varepsilon_{0}A}$. The flux through the area considered is
$\phi_{E} = \frac{Q}{\varepsilon_{0}A}.\frac{A}{2} = \frac{Q}{2\varepsilon_{0}}$
$	herefore$ The displacement current
$i_{d} = \varepsilon_{0}\frac{d\phi_{E}}{dt} = \varepsilon_{0}\left( \frac{1}{2\varepsilon_{0}} ight)\frac{dQ}{dt} = \frac{i}{2}$

Q. 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, the displacement current through this area, will be

Explanation

Reviewed by Dr. Abubacker

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