Electric field in conductor at equilibrium: explanations seem to be lacking

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\sigma$.

PROOF

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.

Answer 4

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.

Answer 5

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.