Let us assume the universe filled with positive charge. About a particular point, all the positive charged particles will be symmetrical. Now consider a sphere of radius $r < \infty$ and apply Gauss law. According to Gauss law the electric flux through the sphere will be some finite value as it contains some positive charge. But by symmetry the electric field through the sphere and hence electric flux will be zero. This is a contradiction.

  • $\begingroup$ any doubt (regarding the question) please ask... $\endgroup$ Jul 15, 2015 at 6:46
  • $\begingroup$ I am not understanding what you are actually saying: Where have you taken the Gaussian surface? You have said the universe is filled with positive charge; so does that mean the surface is enclosing the whole universe?? If that is so, how can it be symmetric????? $\endgroup$
    – user36790
    Jul 15, 2015 at 7:50
  • $\begingroup$ It's like what Kevin said, Sanket. For a simpler example, imagine you're in a spaceship flying exactly between two stars that are fairly close to one another. You don't fall towards either of them, and you might say there's no gravitational field there. There has to be a gradient in potential. $\endgroup$ Jul 15, 2015 at 12:14

3 Answers 3


The problem here is that you've failed to specify a boundary condition.

In an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$, the answer is the solution to the set of differential equations $$\nabla \times \mathbf{E} = 0, \quad \nabla \cdot \mathbf{E} = \rho/\epsilon_0.$$ To get a unique solution to a differential equation, you have to specify a boundary condition. Usually, that condition is "the field is zero at infinity".

In this situation, the usual boundary condition doesn't work because the charge distribution is also infinite. To find the electric field, you must specify a boundary condition. Otherwise, the solution is just as ambiguous as trying to solve $F = ma$ without an initial position or velocity.

Once you do that, the symmetry will be broken, making your "$\mathbf{E}$ is symmetric so must be zero" argument fail. For example, one solution is $\mathbf{E}(\mathbf{r}) = kx \hat{i}$ for some value of $k$. You can check it satisfies Gauss's law. The boundary condition is "the field looks like $kx\hat{i}$ at infinity", which is not symmetric.

This issue is subtle: it also comes up when considering gravitational fields in an infinite uniform universe. Newton made the same mistake, thinking that $\mathbf{g}$ had to vanish everywhere by symmetry.

  • 2
    $\begingroup$ There's a version of this problem that goes the other way in Griffiths's E&M text; it basically starts out with the $\mathbf{E}$ you give, asks for the corresponding $\rho$, and then asks the student to comment on why the field points in a particular direction when the charge distribution is uniform. Hilarity ensues. $\endgroup$ Jul 15, 2015 at 20:58
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    $\begingroup$ Yup, I had that problem in mind while writing this answer! $\endgroup$
    – knzhou
    Jul 15, 2015 at 21:03
  • $\begingroup$ I think you have explained well about the "Gauss's Law paradox" and solved that paradox, but did not really answer the Newton's problem (the symmetry argument) - Newton's argument was really just saying: "no unique direction; hence, no force." We don't need to solve any equation at all. $\endgroup$
    – Shing
    Sep 26, 2019 at 11:28
  • $\begingroup$ @Shing The whole point of this answer is that that argument is wrong. $\endgroup$
    – knzhou
    Sep 26, 2019 at 16:19
  • $\begingroup$ This answer show the "Gauss's Law paradox" is wrong (and it's physically absurd anyway). but I don't think this answer shows Newton's wrong: there is no solving equation or whatsoever in the symmetric argument. $\endgroup$
    – Shing
    Sep 26, 2019 at 16:35

The answer to this question completely depends upon how the charges are distributed.Saying that Gauss law holds here is correct but not in all cases. Suppose that the positive charges are to be randomly placed. Such a distribution would have slight non-uniformity, which would develop further with time because positive repels positive. So, a net flux will leave a sphere containing a point of symmetry.

Gauss law is falsified if all the charges are uniformly oriented instead, with no local non unifomity. In this case, all the charges would become the point of symmetry, hence no charge would remain special, hence no electric feild.

Let me modify this situation for simplicity. Let us assume a uniform positively charged medium spreading uniformly throughout the universe. Since we now assume the whole of universe to be uniformly charged, this is the same as no such charge anywhere in the universe at all.

Note that charging a uniform infinite medium uniformly is like not charging it at all!

So, a universe with a uniform positive charge distribution is physically same as a universe with no such charge distribution at all. So Gauss law in the first universe appears to be violated, while in the second equivalent universe, it is not...


This is a good question because you are about to learn something. As outside charge create electric fields going in, it then also contributes fields going out on the opposite side of the surface. So they cancel themselves out. Where as the charge inside the surface only has electric fields going out.


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