Displacement current, \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \), can also be reached directly from the charge that piles up on a capacitor plate, using Gauss's law, rather than from Ampere's law.
Step 1: Charge on a parallel-plate capacitor from Gauss's law.
For two oppositely charged parallel plates with a uniform field \( E \) between them, Gauss's law applied to a pillbox enclosing part of one plate gives \[E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}\] so the charge on the plate is \[Q = \varepsilon_0 E A = \varepsilon_0 \Phi_E\] where \( \Phi_E = EA \) is the electric flux between the plates.
Step 2: Differentiate with respect to time.
As the capacitor charges, \( Q \) grows with time, and the rate at which it grows is exactly the conduction current flowing in the connecting wire: \[i_c = \frac{dQ}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt}\]
Step 3: Interpret the right-hand side as a current in the gap.
No actual charge crosses the gap between the plates, but the quantity \( \varepsilon_0 \frac{d\Phi_E}{dt} \), calculated purely from the changing field, has the same value and the same units as the conduction current feeding the plate. Maxwell identified this quantity as the displacement current, \[i_d = \varepsilon_0 \frac{d\Phi_E}{dt}\] so that \( i_c = i_d \) at every instant while the capacitor is charging, keeping current continuous through a circuit that otherwise has a physical break at the capacitor gap.
Step 4: Constant voltage case.
Once a constant voltage is applied across a conductor, or once a capacitor has finished charging and reached a steady applied voltage, the charge \( Q \) on the plate stays fixed, so \( \Phi_E \) stops changing: \[\frac{d\Phi_E}{dt} = 0 \quad \Rightarrow \quad i_d = 0\] There is no growing field left to sustain a displacement current, only whatever steady conduction current the circuit allows.