4
$\begingroup$

I have ran into a problem while trying to prove that the electric field is zero in a perfect conductor

My argument went something like this:

We know that: $$\vec J = \sigma \vec E$$

In a perfect conductor $\sigma = \infty$

Therefore to maintain a constant current, $\vec E$ must be zero.

While this $\infty \times 0$ is a constant argument is not never seen before, I feel ike this can be made much rigorous.

Can someone help provide me with an argument why the electric field must be zero in a perfect conductor?

$\endgroup$
  • $\begingroup$ You can define a perfect conductor to be the limiting case as $\sigma \rightarrow \infty$. Then you have $\vec{E} = \lim_{\sigma\rightarrow\infty} \frac{1}{\sigma}\vec{J} = 0$ $\endgroup$ – By Symmetry Nov 20 '14 at 0:54
  • 1
    $\begingroup$ Sorry could you also explain why J must be constant? I'm currently looking into wherever I got this assumption from $\endgroup$ – Carlos - the Mongoose - Danger Nov 20 '14 at 1:01
  • 1
    $\begingroup$ In a sense, a perfect conductor is analogous to a frictionless surface; for a mass with constant speed on this surface, the applied force is zero regardless of the speed. Only in the case of changing speed is the force non-zero. Have you looked at electrical-mechanical analogies? lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html $\endgroup$ – Alfred Centauri Nov 20 '14 at 1:17
  • 1
    $\begingroup$ Firstly $\vec{J}$ doesn't have to be constant, it just has to remain finite (or even just grow slowly compared to $\sigma$) Having infinite current densities doesn't give us physically reasonable solutions, so it is simply more useful to take the limit in this way $\endgroup$ – By Symmetry Nov 20 '14 at 1:19
  • $\begingroup$ @AlfredCentauri: I think irrespective of whether the conductor is perfect or not, it must be possible to prove by vector addition that the field inside the conductor must be zero, since even with a non-ideal conductor vector addition does not pose a limitation. $\endgroup$ – Gaurav Nov 20 '14 at 13:31
4
$\begingroup$

Suppose we impose a current density $\newcommand{\j}{\mathbf{J}}\j$, then the resulting electric field $\newcommand{\e}{\mathbf{E}}\e$ is given by $\e = \rho \j$, where $\rho$ is the resistivity. In a perfect conductor, $\rho=0$. So in a perfect conductor with some fixed current $\j$, the electric field satisfies $\e = \rho \j = 0 \j = \mathbf{0}$. I don't know if this is any more satisfying, but it doesn't use infinity.

$\endgroup$
  • $\begingroup$ I think you're indeed really close : it's $D$ which is zero, not $E$, saying $\rho=0$ is kind of the same as saying $D=0$, since $\nabla\cdot D = \rho $ (Maxwell-Gauß). The confusion is unfortunately common, but it's clear : in a perfect conductor there is no delayed charges, *i.e.* no friction. Most of the lectures discuss the conductor (in fact, circuits) as having only free charges, and so $E\sim D$, then you need the Ohm's law $J=\sigma E$ as a constitutive relation, but it's fake in a perfect conductor, since $\sigma$ is undefined. In a superconductor only $B=0$. $\endgroup$ – FraSchelle Nov 20 '14 at 23:48
  • 1
    $\begingroup$ @FraSchelle, note that $\rho$ is the resistivity, not the charge density. So the equation $\mathbf E = \rho\mathbf J$ is just another way of writing Ohm's law, not a reference to Gauss' law. $\endgroup$ – jabirali Dec 10 '14 at 20:53
1
$\begingroup$

The electrical field $\mathbf{E}$ is an external field, which "drags" the conductor electrons through the conductor "lattice". The conductivity $\sigma$ describes the resistance of the "lattice". When the resistance is zero, a non zero current can exist in the conductor without necessity to support it with an external field.

If $\mathbf{E}$ is non zero, the current will be growing (or varying), just like a solution to $m\mathbf{a}=\mathbf{F}_{ext}$.

$\endgroup$
  • $\begingroup$ Mr. @VladimirKalitvianski, don't you think the OP specifically asked for a 'proof' of the field being zero ? $\endgroup$ – Gaurav Nov 20 '14 at 13:36
  • $\begingroup$ Yes, but it is clear that the macroscopic field can only be zero in case of a constant current $I$. (The current density $\mathbf{J}$ may vary in space, along the wire with different cross sections, like a profiled pipe.) $\endgroup$ – Vladimir Kalitvianski Nov 20 '14 at 13:54
1
$\begingroup$

Can someone help provide me with an argument why the electric field must be zero in a perfect conductor?

It's not clear exactly what you're looking for. In a sense, any argument attempting to prove that the electric field must be zero in a perfect conductor will beg the question.

For example, here's an excerpt from "Electromagnetics for High-Speed Analog and Digital Communication Circuits":

We could in fact define a perfect conductor as a material with zero electric field inside the material. This is an alternative way to define a perfect conductor without making any assumptions about conductivity.

From this starting point, one reasons that

(1) if there's an electric field inside, it's not a perfect conductor

(2) if the material has a finite conductivity and there is a steady current through, there is an electric field inside proportional to the current density

(3) thus, if the material has finite conductivity, it's not a perfect conductor

Again, it's not clear to me precisely what you're looking for. If the above fails to address your question, please update and clarify your question.

$\endgroup$
0
$\begingroup$

Consider a metal sheet placed in a uniform electrical field normal to the sheet surface. The electrons will be "dragged" by the electrical field and form an excess of negative charges on one side of the metal sheet and an excess of positive charges on the other side of the metal sheet. The excess charges produce an electrical field that cancels the applied field in the interior of the metal sheet. The same consideration applies for a metallic half space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.