A parallel - plate capacitor is being charged.Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires.

Explanation

Let `A` is the area of plates , `q` is the charge on capacito r at
some instant and the separation between the paltes
Conducting current ,`i_(c) dq / dt`

Electric field between the paltes,
`E = (sigma )/ (epsilon_(0)) = (q//A)/(epsilon_(0)) = (q)/(A epsilon_(0))`
The flux of the elctric field through the given area is
`phi E = EA = ((q)/(A epsilon_(0))) A = q / epsilon_0`
`:. (d phi E)/(dt) = 1/epsilon_0 ((dq)/(dt))`
Displacement current ,
`i_(d) = epsilon_(0)(d phi_(E))/(dt)`
`= epsilon_0 [(1)/(epsilon_(0)).(dq)/(dt)]`
or `i_(d) = (dq)/(dt)`
From Eqs. (i) and (ii), we can see that
`i_(c) = i_(d)` .