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In electromagnetism books, such as Griffiths or the like, when they talk about the properties of conductors in case of electrostatics they say that the electric field inside a conductor is zero.

I have 2 questions:

  1. We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges?

  2. When the textbooks try to show why the electric field inside a conductor is zero they say let us put our conductor in an electric field. What happens then is that there will be an induced surface charge density which consequently induces an electric field within the conductor such that the total electric field within the conductor will be zero. That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic!

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Conductors are defined by the freedom of some of the charges inside to move with little resistance.

So, if there were a non-zero field, what would happen? Answer: some of the free charges move until the field is again zero.

You might be wondering if there are limits to this claim, but a introductory book of that sort is not worrying about extreme situations. In any case, try choosing a simple geometry, make an estimate of the fraction of charges that are free to move and calculate the saturation field.

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  • $\begingroup$ Will electrons in metals be really stationary? at rest ? what about thermal motion? What about quantum mechanics? $\endgroup$ – Revo Mar 24 '12 at 20:06
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    $\begingroup$ What about quantum mechanics? Even very small surface charges are made up of bjillions of electrons, so it's fair to use statistical measures. As for the non-static nature of the transient, well, yes. The transient is not static and you can't perform a full analysis with the tools of electrostatics, but it is also transient. It goes away and you have a static situation until you introduce another perturbation. $\endgroup$ – dmckee♦ Mar 24 '12 at 20:10
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In electrostatics free charges in a good conductor reside only on the surface.

There are at least two ways to understand this

  • So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is UNIFORM, the electric fields cancel each other.

  • Consider a Gaussian surface inside the conductor. Charge enclosed by it is zero (charge resides only on surface). Therefore electric flux =0 Furthermore, electric flux = electric field * area. Since area cannot be zero, electric field is zero

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    $\begingroup$ Moreover, electric fiels cannot penetrate through a conductor as found in faraday's ice pail experiment $\endgroup$ – Aadhil Azeez Apr 5 '13 at 13:31
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    $\begingroup$ The electric field and "area" are vectors, which can cancel out (for instance, if there is a uniform electric field and you choose a region without any charge in it - then the flux will be zero, but certainly there will be a non-zero electric field present). So how is that proving that the field is zero? $\endgroup$ – Max Mar 27 '17 at 18:23
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Gauss's law states that the electric field flux through a closed surface is equal to the quotient of the load inside the surface divided by $ \epsilon_0$.

A driver is characterized by the charge carriers can move freely within it. If the charges in a conductor in equilibrium at rest, the electric field intensity in all interior points of the same must be zero, otherwise, would move the loads caused an electric current.

Within a conductor arbitrarily draw a closed surface $S$, and it follows that:

The electric field is zero, $E = 0$ on all points of said surface.

The flow through the closed surface $S$ is zero.

The net charge q on the inside of said surface is zero. As the closed surface S we can make it as small as we conclude that at any point P inside a conductor there is no excess burden, so this should be placed on the surface of the conductor. this should answer your question.

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protected by Qmechanic♦ Apr 5 '13 at 20:15

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