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 $\frac{A}{3}$ parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is: |
I $\frac{I}{3}$ $\frac{2I}{3}$ $\frac{I}{6}$ |
$\frac{I}{3}$ |
The displacement current = conduction current = I corresponding to area A. Therefore displacement current for area $\frac{A}{3}$ is $\frac{I}{3}$. |