Consider that space is uniformly charged everywhere, i.e., filled with a uniform charge distribution, $\rho$, everywhere.
By symmetry, the electric field is zero everywhere. (If I take any point in space and try to find the electric field at this point, there will always be equal contributions from volume charge elements around that point that will vectorially add up to zero).
Consequently, from Gauss' law in the differential form $$\nabla\cdot E = \frac{\rho}{\epsilon_0}$$
if $E$ is zero, the divergence is zero hence the charge density is zero.
What is going on here? is a nonzero uniform charge distribution that exists everywhere has no effect and is equivalent to no charge at all?
Best Answer
If you followed the arguments carefully and checked what is demonstrably right and what is not, you would agree that what the argument actually does is to prove that a uniform electric charge density cannot have a uniform electric field. Your original task was to solve Maxwell's equations (well, Gauss's law), so if you find out that the equations aren't satisfied, it just means that you haven't solved the problem you wanted to solve, or that the candidate solution is wrong. You can't suddenly say – as your question suggests – that it doesn't matter that the equations aren't solved and you want to change them or something else. This would be changing the rules of the game – and changing the laws of physics.
Instead, you will be able to find solutions $\vec E(x,y,z)$ which obey ${\rm div}\,\vec E = \rho/\epsilon_0$. However, it is not true that this $\vec E(x,y,z)$ may be translationally symmetric. Instead, you must pick an origin where $\vec E = 0$, let's say at $(x,y,z)=(0,0,0)$, and write $$ \vec E = \frac{\rho}{\epsilon_0} (x/3, y/3, z/3) $$ Feel free to check that the divergence is what you wanted it to be. You may also write this $\vec E$ from a potential, $\vec E = -\nabla \phi$, $\phi = \rho/(2\epsilon_0) (x^2+y^2+z^3)/3 $. I could actually divide the terms asymmetrically to the coordinates $x,y,z$.
The same "paradox" arises in the case of the gravitational acceleration and mass density. In the non-relativistic case, this surprising non-uniformity of $\vec E$, while not paradox, strongly suggests that at the longest scale, the charge density should be zero. And it's indeed the case of the electric charge density. For the mass density, it's not the case although the Newtonian argument would still lead us to conclude that it should be true. However, the mass can't be negative and the energy density is positive. This would force a violation of the translational symmetry in a uniform Newtonian Universe. However, the Einsteinian (well, FRW) universe obeying the laws of general relativity has no problem with an overall positive mass density: the Universe just gets curved accordingly. Our Universe is an example.