# 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.

• What about electric field on the flat surface outside the wire? The capacitor produces some electric field outside the capacitor since there is positive charge one plate and negative charges on the other plate. Commented Dec 7, 2020 at 20:50
• @JánLalinský If the plates are very large and very close together, the electric field anywhere outside the inward-facing surfaces of the plates is negligible (especially if you stay close to the wire and away from the edges of the plates). The field from the negative plate cancels with the field from the positive plate there (at least, to the extent that the approximations are good).
– pwf
Commented Dec 8, 2020 at 0:39

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
Commented Jul 23, 2015 at 18:04

Suppose a capacitor with capacitance $$C$$ and $$d$$ of separation of plates, with the wires being connected by a resistor $$R$$ with length $$L$$ and cross section $$S$$.

Before closing the circuit, there is a voltage $$V$$ in the capacitor.

After the circuit is closed the voltage falls exponentially:$$V = V_0e^{\frac{-t}{RC}} \implies \frac{\partial V}{\partial t} = -\frac{V_0}{RC}e^{\frac{-t}{RC}}$$ As we want the electric field, we divide by the length in each case: When $$t=0$$, we have for the capacitor: $$E = \frac{V_0}{d}\implies \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0}{dRC}$$

In order to calculate the integral over the plate of the capacitor, it is necessary to multiply by its area. $$A = \frac{Cd}{\epsilon_0} \implies \int_A \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0}{dRC}*\frac{Cd}{\epsilon_0} = \frac{V_0}{R} = I_0$$ as expected.

Applying the same integral to the resistor, the voltage have to be divided by $$L$$, and the integral is now over its cross section $$S$$. The term that multiplies $$I_0$$ shows up as the 'capacitance' of the resistor. It is expected to be several orders of magnitude below that of the capacitor (comparing the usual lengths and areas), and this term can reasonably be disregarded.

$$\int_S \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0S}{LRC} = \frac{C'}{C}I_0$$

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.

• Are you saying that the field inside the wire is 0 or that it is nonzero but its contribution is equal to the current I? Commented Feb 15, 2021 at 16:03