Show that displacement current is equal to conduction during charging of a capacitor.

Explanation

Let at an instant magnitude of charge on the plates of capacitor be Q. Area of each plate is A.
Electric field between the plates of capacitor
`E = (Q)/(A_(E_0))" "…(i)`
Flux of this field passing through the surface between the plates is
`phi_(E) = E xx A = (Q)/(A_(epsilon_0)) xx A`
`phi_(E) = (Q)/(epsilon_0)" "(ii)`
Displacement current `l_(d)` is
`l_(d) = epsilon_(0) xx (d phi E)/(dt) = epsilon_(0) ((Q)/(epsilon_(0)))= epsilon_(0) xx (1)/(epsilon_(0)) (dQ)/(dt)`
`l_(d) = (dQ)/(dt) " "...(iii)`
`(dQ)/(dt)` is the rate at which charge is reaching to positive plate of capacitor through conducting wire therefore `l_(c) = (dQ)/(dt) " "...(iv)`
From equations (iii) and (iv)
`l_(d) = l_(c)`