0
$\begingroup$

I am studying electromagnetic waves, and in many texts it is stated that when a plane wave hits the surface of a perfect conductor, the electric field must be zero in the surface and inside the body. Why is that? What happens inside the conductor so as to cancell the field completely and what is the mechanism behind it? It would great that any answer given is within the scope of classic electromagnetism, since I have no knowledge of quantum mechanics.

Edit: This question is in the context of the study of stationary waves, as it is used as a boundary condition that $\vec E=0$ at a specific plane in which physically there is a conductor metalic sheet. What I am specifically looking to know is why when the incident wave acts on the perfect conductor (the book I am reading from refers to perfect conductor a material with zero resistivity) the E field must be zero. How the moving charges inside of the conductor cancel perfectly out the E field and not also the B field, for instance.Moreover, the book also argues that these induced currents generate the reflected wave that in combination with the incident one, creates the stationary wave.

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
2
$\begingroup$

First of, notion of:

$$\lim_{\sigma \to \infty} (\vec{J} = \sigma \vec{E})$$

Doesn't make any physical sense, perfect conductors don't exist, and the mathematical formulation of ohms law doesn't even apply here, since ohms law is the steady state current when acceleration reaches zero. In the absence of resistance, there is no steady state solution, and hence the current is infinite for any electric field.

With that being said, solving maxwells equations with ohms law and taking sigma to infinity yields that the field vanishes.

This is due to the EM field created by the moving charges, the field superposes with the incident wave, cancelling out.

A better way of dealing with similar problems like this, is using the London equations, these model electrons as free electrons, which is the best kind of "conductivity" that you will get.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Ofcourse if there is static E field in the conductor, the steady state of the electrons move to cancel out this external field. $\endgroup$
    – jensen paull
    Commented Nov 30, 2022 at 19:27
  • $\begingroup$ Doesn't the same cancelation apply to a non static E field at any instant in time? $\endgroup$
    – Bob D
    Commented Nov 30, 2022 at 19:48
  • $\begingroup$ If it did at every instant, there would be no transient stage to move the electrons to a configuration to cancel the external field. Eddy currents would not be generated to produce the meissner effect. There must be atleast some time t, at which there is an electric field, otherwise nothing would happen. Although I guess at some level there would also be a steady state in the time dependant case(after some time t), however wouldn't cancel the field completely, atleast up until the London penitration depth $\endgroup$
    – jensen paull
    Commented Nov 30, 2022 at 19:53
  • $\begingroup$ Considering the edit I just added to my question, if you do not mind answering: How does the EM field generated by the induced currents in the conductor cancel out perfectly the E field from the wave? Is it beacause they acumulate on one side, creating a counter E field? Also, is the interaction of the wave's B field with the charges considerable? why does the B field inside the conductor survive (I read that in superconductors where rho tends to 0, the B field inside is also 0)? $\endgroup$
    – user336281
    Commented Nov 30, 2022 at 21:23
  • $\begingroup$ The skin depth formula that you see, Are assuming Maxwells equations under the condition that $\vec{J} = \sigma \vec{E}$. This predicts that $\vec{E}=0$ assuming the conductivity everywhere in space is infinity, and that $\nabla \cdot \vec{E} = 0$ Meaning $\rho_{net}= 0$ . The field still vanishes, whilst also obeying the condition that charge does not build up anywhere, so this cannot be a valid explanation. The field of a moving charge is different from the field of a stationary one, it is the field of the moving charges that cancel it. $\endgroup$
    – jensen paull
    Commented Nov 30, 2022 at 21:59
1
$\begingroup$

There is a classical explanation: A conductor has a large conductivity that means it has a large number of the mobile charge carriers (usually electrons).

These charge carriers move easily under the influence of any applied electric fields and their motion nullifies this applied electric field.

Improve this answer
$\endgroup$
0
0
$\begingroup$

Let me describe my biased view.

For metallic antennas driven at frequencies from a few kHz to GHz, the strength of the electric field inside metal is sufficiently small compared to its strength in space. How about visible light frequencies?

Roughly, electromagnetic properties of materials can be divided into two kinds of objects: "Nevertheless, most everyday objects belong (at least, in good approximation) to one of two large classes: conductors and insulators (or dielectrics)." (D.J.Griffiths, Introduction to ElectroDynamics; Chap. 4).

Here I will describe my own biased speculation. Metals in reality behave as conductors at low frequencies and as insulators at high frequencies, the frequencies of visible light. An "insulative metal" in the optical wavelength range is mainly described by the real part of the permitivity (permitivity is, roghly, property of insulator), while losses can be expressed by the imaginary part of the permitivity. A well-known sample model is the Drude model.

For a real metal antenna at low frequencies, I think it is desirable to "almost cancel out the electric field inside the conductor". At high frequencies, there is no conductor material.

Improve this answer
$\endgroup$