# 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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

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.

An electric field crossing the flat surface of the Amperian loop, implies some difference of potentials, or voltage, between the capacitor plates.

However, if the plates are connected by a wire, assuming that this wire is a good conductor and the (conducting) current flowing through the wire is not huge, the voltage drop on the wire, and therefore, the voltage between the plates, will be negligible. Hence, the electrical field crossing the flat surface of the Amperian loop will be negligible as well.

You may ask if the plates of the capacitor will have some charge on them. They may, because, as the current flows around the circuit, some parts of the circuit would acquire a little positive charge and some a little negative charge, but both plates will have pretty much the same charge density (all positive or all negative, depending on the position of the capacitor in the circuit, i.e., closer to the positive terminal of the battery or negative) because they are shorted by the wire and therefore no field would develop between them.