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This question stems from the exercise 2.50 of Griffith's Introduction to Electrodynamics, which I am rereading.

In the problem, he asks to find the charge density of a volume when the electric field is

$\textbf{E}(x,y,z) = ax\space \hat{x}$

And to find a possible explanation as to why the electric field would point in that particular direction.

We can easily show that electric field corresponds to a uniform charge density, but it is not simple to explain why, then, a uniform charge density would give rise to such an electric field.

If we want to answer why the field is oriented along the x axis instead of any other, we can argue that knowing the divergence and rotational of a field is not sufficient to determine it uniquely, and thus, a uniform charge density definitely could give rise to, say, $\textbf{E}(x,y,z) = ay\space \hat{y}$. This ambiguity would be lifted by the appropriate boundary conditions, which is trivial, in a way.

What I cannot grasp at all is how, if we consider a non-zero infinite distribution, these results even make sense. Due to rotational and translational symmetry, an infinite uniform charge density should give rise to no electric field. However, if we calculate the divergence of a null field, we get $\nabla \cdot\textbf{E}(x,y,z) = 0$ instead, which would require $\rho = 0$, in contradiction with our initial assumption.

I have tried to use Gauss's law to analyse the problem, but I reach a similar nonesensical answer: Consider a sphere in this infinite charge distribution, as well a spherical shell that touches the surface of the sphere and has infinite thickness (you can also consider an infinite number of shells with finite thickness that fill the whole space). From Gauss's law, the electric field the shell creates in the surface of the sphere is zero, while the electric field created by the sphere is proportional to its charge. Thus, the electric field in the surface of the sphere is radial and non-zero, in contradiction once again with symmetry arguments.

What is going on here?

Qmechanic
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Andre P.
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  • Also I think there is some stuff in the appendix that night help illuminate this. – BioPhysicist Jul 07 '19 at 23:28
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    Yes, you are correct, I meant $ay \hat{y}$ Can you be more specific? I don't see how the appendix might help. – Andre P. Jul 08 '19 at 00:22
  • Helmholtz Theorem. You seem to dismiss the boundary conditions, but that is actually the key here. – BioPhysicist Jul 08 '19 at 00:28
  • As I said in the text, I understand about the boundary conditions and their role in lifting the degeneracy of solutions, so to speak, but I do not see how they are relevant in an inifinite distribution, since all space is uniform and we effectively have no boundary. Gauss's law uses no boundary conditions either, and the contradiction still arises. – Andre P. Jul 08 '19 at 00:34
  • You have your answer right there. That's the point. You can't define natural boundary conditions here. – BioPhysicist Jul 08 '19 at 00:40
  • You can't define them, but you still get two different answers using symmetry arguments and Gauss's Law. Why? If we define the boundary conditions of said sphere to be electric field of 0, from symmetry, but then apply Gauss's law, don't we still get a contradiction? – Andre P. Jul 08 '19 at 00:44
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    Possible duplicate of Gauss's law in a uniform charge distribution extending infinitely in all directions – knzhou Jul 08 '19 at 01:02
  • @knzhou Ah I'm usually good at looking for duplicates first. Good catch. – BioPhysicist Jul 08 '19 at 01:05
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    @knzhou Thank you for that link. I did a search before posting, but for some reason I didn't find that post. Probably didn't write the correct query. – Andre P. Jul 08 '19 at 01:19
  • So, if I get correctly, when we employ Gauss's law naïvely, we are assuming that the field is zero at infinity, which works as a boundary condition? If that is so, then how does that work for an infinite charged plane? Gauss's law still works, but the field is not zero at infinity, and the symmetry around the plane shouldn't be enough to lift the indetermination. – Andre P. Jul 08 '19 at 01:24
  • Have you read that part of the appendix yet? – BioPhysicist Jul 08 '19 at 05:23

1 Answers1

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The issue is that you can't even talk about the electric field caused by a uniform charge density that fills all space. In other words, the field $\mathbf E=ax\ \hat x$ in all space just doesn't work.

As you hint to in your answer, the divergence and curl of a field does not uniquely determine the field. We need appropriate boundary conditions. Due to Helmholtz's Theorem, we need the field to drop off fast enough as we go to infinity to determine the field. However, for a uniform charge density that fills all space we can't have this condition. We also can't think of any smart symmetries to help us (which direction would such a field point?).

Therefore, any arguments you try to make further with this field are invalid. I don't think I fully understand what you are trying to say in your last paragraph about using Gauss's law, but this is probably what the issue is. You are working in a system where we shouldn't actually be able to define a unique field (as you noted with your $ay\ \hat y$ example). It should also be noted that $\nabla\cdot\mathbf E=\rho/\epsilon_0$ is also Gauss's law, so if that is "broken" in this case then arguments using the integral form should also be "broken".

BioPhysicist
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  • I am sorry, but I still don't understand your explanation. When I applied Gauss's law to a point charge, line charge, or plane charge, I do not recall ever defining boundary conditions. – Andre P. Jul 08 '19 at 01:28
  • @AndreP. The necessary boundary conditions and/or symmetries are possible in that problem to make the problem doable. In this problem that isn't the case. You can use Gauss's law to determine the field in those situations because of the boundary conditions. You usually just don't check that the appropriate boundary conditions can be applied because we usually don't encounter situations like this where they don't hold. – BioPhysicist Jul 08 '19 at 01:37