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//4` parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Explanation

Let at some instant change on capacitor be `Q`. Electric field between plates of capacitor
`E = (Q)/(Ain_(0))`
Flux passing through area `A//4`
`phi_(E) = E.(A)/(4) = (Q)/(A in_(0)).(A)/(4) = (Q)/(4 in_(0))`
The displacement current
`i_(d) = in_(0)(dphi_(E))/(dt) = in_(0)(1)/(4)(dQ)/(dt) = (1)/(4)i`