# In terms of the Ampere-Maxwell law, why is $\vec {E}=0$ in a wire of a capacitor circuit?

I'm currently studying from "Introduction to Electromagnetics" by D.J. Griffiths. In the book the significance of the displacement current term is explained by looking a non-steady capacitor circuit (shown below).

Using the integral equation and balloon-shaped surface, it is said that $I_{enc}=0$ but $\int \partial\vec{E}/\partial t \cdot d\vec{a}=I/\epsilon_0$. This statement makes sense to me - there is no physical current flowing between the plates but there is a changing electric flux through the balloon surface.

My misunderstanding is with the flat Amperian loop surface where it is mentioned that $\vec{E}=0$ and $I_{enc}=I$ in this case. Obviously there is a current flow $I_{enc}$ in the Amperian loop, but why is $\vec{E}=0$? Just looking at the equations directly, I would have thought there would be a combination of both current terms (including displacement current due to changing electric field in the wire) in a capacitor charging/discharging situation.

I've attempted to seek other explanations on this from other texts, this forum and elsewhere in this stackexchange site, but this has made me more confused. Clarification on this would be appreciated.

Differential form of Ampere's-Maxwell's equation: $$\nabla \times \vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$$

Integral form of Ampere's-Maxwell's equation: $$\oint\vec{B}\cdot d\vec{l} = \mu_0I_{enc}+\mu_0\epsilon_0 \int \frac{\partial \vec{E}}{\partial t} \cdot d\vec{a}$$

The derivation assumes the wire is a perfect conductor, and also that it is negligibly thin. If it had some resistivity, then you're right, there would be an electric field in the wire, but even in that case the electric flux $\int \vec{E}\cdot\text{d}\vec{a}$ would be negligible, and so would its time derivative.

There is never actually an electric field in a conductor in the electrostatic sense. An E field is always generated perpendicular to a charged surface (the wire). For any wire carrying current, the electric field tends to radiate outward from the wire. The magnetic field will be circulating around the wire such that the Poynting vector, $\vec S = \vec E \times \vec H$ , is pointing in the direction of current flow, which is also the direction of power transfer. So for your flat Amperian loop, the E field is parallel to the radius of the loop, so there is no net electric flux.

I believe if you were operating at high enough frequencies so that there would be a non-negligible E field inside the conductor, this circuit model really wouldn't be applicable anyway.

• There would be no radial electric field if the wire were neutral, i.e. if the current were composed of charges of opposite sign moving at different speeds. This is usually the case in circuits; electrons flow, but their charge is balanced by stationary positive ions in the material. – pwf Jul 23 '15 at 18:04

For the flat Amperian Loop,

The current flowing through the wire that pierces the surface of the loop) is I. However, there is no field piercing the surface of the loop.

Now, why is there is no field piercing the loop:-

1. Of course the field between the plates of the capacitor no way pierces the surface of the loop

2. "Isn't there a field inside the wire, which is piercing the loop surface?" The answer is no. Note that you are already accounting for this field by taking the current I into the calculation and hence there is no need to consider again the field inside the wire.