Consider a parallel plate capacitor is charged by a constant current I, with plate area A and separation between the plates d. Consider a plane surface of area `A//4` parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Explanation

Charge on the capacitor plates at time t is given by,
`q = It`
Electric field between the plates of the capacitor at this instant is,
`E = (q)/(Aepsi_(0)) = (It)/(Aepsi_(0))`
So, elecric flux though the given area,
`phi_(E) = ((A)/(4))*E = (It)/(4epsi_(0))`
Therefore, displacement current
`I_(d) = epsi_(0) (dphi_(E))/(dt) = epsi_(0) (d)/(dt) ((It)/(4epsi_(0))) = I/4`