- #1

Hakkinen

- 42

- 0

Basically the "usual" symmetry arguments used for examples like an infinite line/plane of charge when first applied to this problem seem to suggest that the field anywhere should be 0. However then it's straightforward to see from the differential form of Gauss's Law

[itex]\nabla\cdot \vec{E}=\frac{\rho }{\varepsilon _{0}}[/itex]

that the left hand side is now zero. So now you have zero being proportional to the constant charge density rho!

The first thing I thought about this result is that Gauss's Law is saying you simply can't have a nonzero uniform charge density in an infinite volume but it's clearly not all there is to it.

I'm almost certain that the problem lies not in Gauss's Law but in the symmetry arguments used or in some subtlety of the way the problem is described. Once it can be shown that E=/= 0 then the problem should be resolved.

I've read some discussions online about this and similar types of problems and most of the suggestions are something like these:

**1**

In all of the lower dimensional cases of an infinite charge density, the object was placed in 3d space so there were dimensions where the electric field could permeate and be nonzero. So could the problem be that we are viewing a 3d object in 3d space and if viewed in a 4d space (what would that even entail? surely it wouldn't a time dimension and I can't think of reasons to simply add a 4th spatial dimension) then there would be another dimension for the electric field to have a nonzero component.

**2**

It's actually a problem with the divergence theorem since it requires vector fields to be compactly supported, meaning they go to zero at some finite value. So in this case with an infinite volume the field would not be compactly supported and you cannot use Gauss's Law.

If this, #2, is the case then why do the examples of an infinite line/plane charge have no similar paradox like this example?

**3**

You can treat the infinite volume as an infinite sum of infinite plane charges but it might be that the integral is not absolutely convergent, that is, the electric field you get will depend on the order of which you summed the contributions from each plane.