# No electric field inside a conductor by Gauss's Law

So recently in our electrostatics course the lecturer keeps repeating a certain set of steps whenever he solves a problem, and I am not quite sure why these assessments are true.

1. First, he makes the assumption that the conductor is ideal, and that hence all charge resides on the surface.
• We've been only doing spheres, planes and cables, but does that actually apply to every conductor? And why?
2. Hence if we put a Gaussian surface inside the conductor, the surface encloses no charge. By Gauss's Law, that means there is no electric flux (and hence field) inside a conductor.
• This seems to me like the lecturer jumped a few steps there. From my understanding of Gauss's law, it says that the net electric field through the surface is zero, i.e. there could be electric field at any point on or within the surface, but in some places it points towards the inside of the surface, and towards the outside in others. Hence, when you integrate over the surface, the result will be zero, but the field at any given point is not necessarily zero.
• In fact, when I tried calculating electric field inside an infinitely long wire with charge uniformly distributed on it's surface by using Coulomb's Law, it seemed rather obvious that the field does not equal to zero anywhere except for the centre of the wire.

Is there something I am missing here? Some additional implied assumptions? Or is the lecturer overgeneralising?

See, here the reasoning is such that you go from the fact that there is no charge inside the conductor to using Gauss's law to state that the electric field inside the conductor is 0 everywhere. However, this is faulty. The very premise of your reasoning should be that there is no electric field inside the conductor. Think about this, if there is an electric field inside the field then the free electrons of the conductor will start moving and a current will be created although there is no voltage applied. This is impossible and hence E=0 everywhere inside the conductor.

Now, use Gauss's law to get the fact that there can be no charge inside the conductor as any closed surface inside the conductor will have zero flux coming out of it( No electric field linked with the surface area). Hence, any charge provided to the conductor must reside on the surface. This is the simplest possible explanation.

I can not see how you used Coulomb's law to get a field inside the wire( Remember, Gauss's law is a far more fundamental law than Coulomb's law).This should hold for a really long wire as well as it is a conductor as well as long as you do not apply potential difference across its ends. Apply the same logic as above. It should be easy to see the truth.

• It makes sense, but when you say that the electric field inside a conductor must be zero, does that really apply to all cases? Because if the charge already resides on the surface of the conductor, there could very well be an electric field within the conductor, but there would be no charges it could act upon, hence no charge movement. Feb 4 '17 at 18:34
• No, it does not matter whether there are charges are not. There are always free electrons in the conductor. In presence of an electric field, the electrons will start moving and constitute a current without any voltage to sustain it. Feb 4 '17 at 18:37
• @TheFool "Remember, Glass's law is a far more fundamental law than Coulomb's law" Have a look at physics.stackexchange.com/questions/167786/… Feb 4 '17 at 19:14
1. Yes this is always true, and this is related to 2.

2. The key point here is to find a Gaussian surface on which, for reasons of symmetry, we know the field must be constant in magnitude so that, in the flux integral, one can factor" the magnitude of the field $$\oint \vec E\cdot d\vec S = \vert \vec E\vert \oint dS$$ This usually requires not only a constant magnitude but also a constant dot product $\vec E\cdot d\vec S$: notice how, on the right hand side, the surface element is no longer a vector.

By Gauss's law, the flux is $q_{encl}/\epsilon$, so if you already argued that the field must have constant magnitude on the surface, the is must be that $$\vert \vec E\vert S = 0$$ if there is no net charge enclosed.

If the field is NOT constant on the Gaussian surface - for instance imagine a box where, in one corner there is a positive charge an in the other a negative charge, then nothing can be said about the electric field since $\oint \vec E\cdot d\vec S \ne \vert \vec E\vert \oint dS$. Even if you know the net enclosed charged, you cannot recover $\vert E$ since it is not constant.

Not, inside a conductor, there is no charge irrespective of the shape of the Gaussian surface. This this is the argument "in reverse", i.e. $\oint \vec E\cdot d\vec S=0$ always, irrespective of the shape of the surface, which implies $\vec E=0$ for this to hold.

In the case of your infinitely long wire, it is complicated to use Coulomb's law (because of the geometry of the system, especially if you want the field away from the axis). Then a Gaussian cylinder coaxial with the axis of your wire will not enclose any charge, but the field is constant on the surface of this cylinder. Thus the field will be $0$ everywhere inside.

As an intuitive explanation, consider the following picture:

You see that the point is off-center. The charges of your cylinder on the part of the wall closer to the point are closer but they are geometrically balanced by the charges on the opposite side of the cylinder , which are farther from the point but greater in number. Geometrically the increase in the number of charges exactly balances the decrease in the electric field generated by these charges.

Note that the red circle is not a cross-section of the Gaussian cylinder, which would be coaxial with the cylinder, but is there to illustrate how the flux through both angular openings are the same.