# Displacement current in charging capacitor [duplicate]

In the classic example of a charging capacitor I can see how an Amperian loop with a surface going between the capacitor's plates would feel a changing electric flux, as the electric field between the plates changes with time.

However, considering the same Amperian loop but with a flat surface crossed by the wire leading to the capacitor, there is no displacement current, only conducting current.

The fact that there is no displacement current tell us that there is no change in the electric field across that flat surface.

Can someone elaborate more to me why there is no changing electric field across the flat surface too? I can't understand it...

## marked as duplicate by Rob Jeffries, knzhou, ZeroTheHero, Kyle Kanos, Michael SeifertApr 25 '18 at 16:43

When you consider a loop encircling the wire, you have an electrical field only in the wire. And in the wire this field produces exactly the same total (mainly conduction) current as the displacement current in the capacitor. If you want to be nitpicky, you also have displacement current in the wire when there is a changing electric field. In general, total current $I_{tot}$ continuity holds, which means that along a current flux tube (current loop) the sum of conduction and displacement current is constant $$I_{tot}=I_{con}+I_{dis}$$ This follows from taking the divergence of the Faraday-Maxwell equation, which gives $$div (\vec j+\epsilon \frac {\partial \vec E}{\partial t})=0$$ which means that the total current must be continuous.