I'm trying to understand why the field is zero. This is not a duplicate of previous similar questions because I'm questioning the standard explanations.

I've seen two explanations:

  1. Gauss's law.
  2. A nonzero field would contradict equilibrium.

First, the Gauss Law explanation. The reasoning goes that the charge migrates to the conductor's surface, and thus a Gauss surface in the interior encloses zero charge, so the flux through the Gauss surface must be $0$, and therefore he field must be $0$ too. This doesn't seem right for two reasons. First, why must the charge be located only on the surface? (In fact, it's usually explained the other way around: once we establish that the field is $0$ in the conductor's interior, Guass's law immediately implies that no charge can be present there.) One might argue that the charges repel each other and therefore end up on the surface, although what's to prevent some point charge from being equally repelled from all directions and thus remaining in the interior (e.g., a point charge at the center of a ball). But let's assume we can prove independently that all the charge must be on the conductor's surface. The problem remains that although Gauss's law then implies that the total flux through a Gauss surface in the interior is $0$, this in no way implies that the field is $0$, only that the total flux entering the surface equals that exiting it. (To drive the point home, consider a point charge and a gauss surface, say a sphere, next to it (but not containing it). There is no charge within the sphere. Does that mean the field there is $0$ ?) Typical applications of Gauss's law usually invoke symmetry to infer that the electric field is uniform over the Gauss surface, so it can be factored out of the integral, and in that case $\textrm{flux}=0$ indeed implies $\textrm{field}=0$. However, for a non symmetric conductor this argument breaks.

Turning to the equilibrium argument, the only thing that can be inferred from equilibrium is that if there is charge in the conductor's interior, then the field there must be $0$, but in fact we know there is no charge there, so how can a nonzero field there contradict equilibrium? The equilibrium argument works at the conductor's surface to show that the field there must be perpendicular to the surface, but I don't see how it applies to the conductor's interior, which is charge free.

So I guess my question is: can the statement that the electric field is zero inside a charged conductor be proven solely from Coulomb's law?

  • 1
    $\begingroup$ Just by the by, Coulomb's Law comes from Gauss' Law. $\endgroup$ – user36790 Oct 26 '16 at 7:33
  • $\begingroup$ It sure can be derived from it, and (not being a physicist) I won't argue which of them is more fundamental, but that's not the way textbooks teach it. $\endgroup$ – Ari Oct 26 '16 at 7:58

the only thing that can be inferred from equilibrium is that if there is charge in the conductor's interior, then the field there must be 0, but in fact we know there is no charge there, so how can a nonzero field there contradict equilibrium?

You may be confusing the assumption of zero net charge and zero charge.

The interior of the conductor consists of many fixed charges (atomic nuclei and most of the electrons) and many mobile charges (the valence electrons). Normally these are paired so that the total net charge is zero. When discussing the charge on a conductor ($Q$), we mean the total net charge.

But an electric field doesn't affect only the net charge, it creates forces on all charges. Inside a conductor, this means that an electric field would accelerate the mobile charges. The interior is not charge free.

  • $\begingroup$ Well, the model of an idealized conductor is (to my understanding) a body in which point charges may move freely, and these are the only charges to be reckoned with. If you add to the model fixed protons and mobile electrons, then equilibrium could be achieved even with a nonzero field in some point or region in the interior, provided that the electrons have all been evacuated from there. The protons would constitute a non-zero net charge, and that wouldn't contradict equilibrium because they are fixed. $\endgroup$ – Ari Mar 12 at 14:03

why must the charge be located only on the surface?

Only net charge density must be zero inside, not density of the mobile charge carriers(in metals, electrons). There are still charge carriers inside the metal, in fact the inside is electrically in the same state regardless of whether there is some surface charge or not.

The reason net charge density must be zero inside is the Gauss law combined with non-zero conductivity of the metal. Gauss's law requires that if charge density at some point is non-zero, electric field has to vary in space at that point. But then the electric field at some point nearby would have to be non-zero. Non-zero electric field would make the mobile charge carriers move (due to non-zero conductivity), electric current would occur. Such current would move the system to a state closer to the equilibrium state, where net charge density and electric field is zero inside.

Mathematically, this can be described as follows. We can formulate differential equation for charge density in ohmic conductor.

Ohm's law: $$ \mathbf j = \sigma \mathbf E $$

Law of local conservation of charge:

$$ \partial_t \rho = -\nabla\cdot \mathbf j $$

Combining these two, we obtain

$$ \partial_t \rho = - \sigma \nabla \cdot \mathbf E $$

Now we can use the Gauss law $\nabla \cdot \mathbf E = \rho/\epsilon_0$ and obtain

$$ \partial_t \rho = -\frac{\sigma}{\epsilon_0}\rho $$

Solution of this equations is a decaying exponential function of time. Thus if some charge gets implanted inside the metal, its density decays very fast to zero.

  • $\begingroup$ Your answer is convincing if we assume charge density is finite everywhere, but if we allow point charges, then what's to prevent a point charge from remaining in the center of a ball, as I've asked? But, again, my main problem is not with the assumption that the net charge in the interior is 0. As I wrote, even accepting that, how does it follow that the field must be 0? $\endgroup$ – Ari Mar 12 at 14:33
  • $\begingroup$ Point charge of macroscopic charge magnitude would have macroscopic Coulomb electric field around it and this field would act on the mobile charge carriers around it and electric current would occur, which would mean the body is not in equilibrium. This process is described by the equations above. Your main problem is resolved by the same argument as in the first half of my answer: non-zero field would mean electric current, which would contradict equilibrium. $\endgroup$ – Ján Lalinský Mar 12 at 16:13
  • $\begingroup$ Doesn't non-zero field imply current only if there are charged particles present within the region of that non-zero field? How can a non-zero field in the interior of the conductor induce current if all the mobile charge carriers have already rearranged themselves on its surface? $\endgroup$ – Ari Mar 13 at 14:33
  • $\begingroup$ There are always mobile charged particles present in the interior of the conductor. Pulling all of them to the surface would require insane amount of energy and extreme external electric field; the charges would be pulled back by the positive nuclei with extreme forces. What (in statics) usually ends up on the surface is only excess charged particles, and if there are none (neutral body), what happens is only a small part of all available charge rearranges a little. The interiour hardly ever changes in statics; it takes time-varying fields to disturb the concentration there. $\endgroup$ – Ján Lalinský Mar 13 at 19:00
  • $\begingroup$ Right. Let's accept that there are electrons in the interior held in place by the fixed protons. Because they are fixed, even if there were a non-zero field in the interior, the electrons would still stay in place, so the fact that they aren't moving doesn't imply nonzero field. For this reason, the standard model of charged conductor (as I understand it) ignores the issue entirely, and just assumes there is some charge (in reality, the excess) that is free to move within the conductor. Once that charge relocates to the surface, there is no charge in the interior. $\endgroup$ – Ari Mar 14 at 11:29

In short, i can present a proof directly from the l Laplace equation,but using coulombs law, it is impossible to prove that at an interior point in a conductor,the field is zero, without any prior knowledge of the geometry and surface charge distribution $\sigma$.


Consider the arbitrary surface which bounds the conductor to be $S$. Let $V$ be the electrostatic potential function. Note that in the region bounded by $S$, $\nabla ^2 V = 0$at all points inside $S$,because the interior is charge free(this is basically because $\nabla . \vec E = - \nabla ^2 V = q/{\epsilon_0}$.). We also know that the potential at the boundary surface $S$ is constant over it. So $V(S) = C$ for some constant potential C. Now, the Uniqueness theorem in electrostatics tells us that, if there exists some 2 solutions for the potential function, $V_1$ and $V_2$ satisfying the laplace equation, and having $V_1(S)= V_2(S) = C$, then $V_1 = V_2$ for all points, i.e there exists a unique solution given the boundary conditions. Clearly, if we can guess one solution for $V$ satisfying the laplace equation and having $V(S) = C$, then it is guranteed that this $V$ must be the only possible solution. One can trivially see that $V = C$ everywhere inside $S$ satisfies both the laplace equation and the boundary conditions. In other words, the potential at any interior point is the same as the surface potential C. By definition of potential, then $\nabla V = -\vec E = 0$ at all interior points.

NOTE: you can also extend this proof for the case of a conductor with cavities. I sugeest you read up on the uniqueness theorem.

  • $\begingroup$ There are two assumptions here. The first is that the boundary surface is equipotential. Presumably, if it hadn't been, then some charge on the surface would be accelerating. I'm not sure there isn't some weird geometry and charge density distribution in which there would be a charge free area on the surface with a non-constant potential there, but I'm willing to accept that either this is in fact impossible, or is so esoteric as to never occur in practice. The second assumption is that the interior is charge free. Why is that? $\endgroup$ – Ari Oct 27 '16 at 11:50

I think you have to go back to the definition of an idealized conductor. The defining property of a conductor is that there is an unlimited amount of free charge available. I.e. at any point inside the material there are positive and negative charges that are free to move. If there's an electric field present in the interior then the charges that are present at that point will feel forces and move around.

If we have equilibrium then there can't be an internal field. If there were then free charges would move and the system wasn't actually in equilibrium. Jan's answer contains a quantitative description of this fact.

  • $\begingroup$ If I understand your answer correctly, the key is that there is unlimited amount of free charge available. However, if that is the case, then we are not justified in assuming that equilibrium is necessarily reached. Charge (positive and negative cancelling each other) can just keep flowing forever from the interior to the surface. There's an unlimited amount of it, after all. $\endgroup$ – Ari Mar 12 at 14:13
  • $\begingroup$ You're right -- and you don't always have this equilibrium in a conductor. For instance, if you wrap a wire around the conductor and drive an oscillating current through it the changing B fields will certainly induce E fields in the conductor. But if you are in equilibrium then there is no E field in a conductor. $\endgroup$ – Alex Mar 12 at 15:18
  • $\begingroup$ The context of the question is not one in which an external fluctuating field is applied. Let's assume the universe is empty except for the charged conductor. $\endgroup$ – Ari Mar 13 at 14:15
  • $\begingroup$ Then I think it's reasonable that the situation will eventually come to an equilibrium. And once reached, the electric field must be zero inside. $\endgroup$ – Alex Mar 13 at 14:26
  • $\begingroup$ But that's exactly my question. Why does equilibrium on the surface necessarily imply zero field in the interior? $\endgroup$ – Ari Mar 13 at 14:34

By definition, a conductor is a material where electrons can easily move from atom to atom. With this in mind, a conductor must be equipotential because otherwise, the electrons will freely move until the charge density is constant everywhere (basically because all electrons will feel the same force). Since the electric field is the rate of change of the electric potential along a direction, we can conclude that the electric field must be zero inside a conductor.

  • $\begingroup$ You say otherwise, the electrons will freely move until the charge density is constant everywhere but in fact that would not make it equipotential. The claim is actually that they would all move to the surface, and then the interior would be equipotential. But if there are no electrons in the interior, why must it be equipotential. $\endgroup$ – Ari Mar 13 at 14:12

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