Consider a paralled plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.

Explanation

Given: Radius of the plates = R
area of the parallel plate capacitor = A current in the wire connected to the plates `I_(c )` = 5 A
the electric field between the plates E = `(Q)/(epsilon_(0)A)`
`Q rarr` charge accumulated at the positive plate.
the flux of this field through the given area `Phi_(E)`
`Phi_(E) = (Q)/(epsilon_(0)) xx A = (Q)/(epsilon_(0))`
the displacement current `I_(d)` =
`I_(d) = epsilon_(0) cdot (d Phi_(E))/(dt) = epsilon_(0) cdot (d)/(dt)((Q)/(epsilon_(0))) = (dQ)/(dt)`
`(dQ)/(dt)` is the rate at which the charge is carried to the positive plate through the connecting wire
`therefore I_(d) = I_(c ) ` = 5A.