# Gauss law and infinite spherical charge distribution

1. The problem statement
Consider an infinite spherical charge distribution with constant charge density. According to symmetry of the problem, I expect the electric field at any point to be zero. But if you construct a Gaussian sphere and apply Gauss theorem, it will give you some finite field at that point. Is it a paradox or I am missing something here?

## Homework Equations

$$\nabla E=\rho/\epsilon_0$$

## The Attempt at a Solution

Is there any problem in the symmetry argument or an infinite charge distribution is the problem? I don't think that Earnshaw's theorem should be considered here as the charge distribution can be fixed using other forces (theoretically) and the Gauss' law will still hold.

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## Answers and Replies

Similar (or may be dissimilar) paradox:
The electric field due to an infinite sheet of charge (constant surface charge density) is constant, away from the sheet. But what about the electric field at a point "on" the sheet. According to symmetry, there should be no electric field. However as soon as you move away from the sheet even infinitesimally, there will a constant electric field, independent of the distance from the sheet.

Anybody please.

ehild
Homework Helper
The divergence of $$\vec E = a\vec r +\vec b$$ is constant.

ehild

Let's have a point of coordinates A around which you define a sphere of radius R.
For any point P on the surface of the sphere, you can define another sphere of radius R whose center is at coordinates (2P-A).
P is on the surface of these two sphere.
Based on your assumption we have that the electric field vectors of these 2 spheres in P are equal and opposite, so the net field is zero.
This holds for any A, any P.

@Quinzio
For the sphere with center A and radius R, contribution of the charge distribution outside this sphere will be zero. Due to the remaining spherical charge distribution, electric field will be zero at the center A while non-zero at other points. This is the paradox. What you have given is trying to somehow prove that the electric field will be zero. Choice of Gaussian surface does not change the physics of the problem, it can change its mathematics though.

ideasrule
Homework Helper
1. The problem statement
Consider an infinite spherical charge distribution with constant charge density. According to symmetry of the problem, I expect the electric field at any point to be zero. But if you construct a Gaussian sphere and apply Gauss theorem, it will give you some finite field at that point. Is it a paradox or I am missing something here?

It's physically impossible to have an infinite spherical distribution, so let's consider what happens with a finite distribution. Suppose I have a very large sphere with a certain charge density, rho. At the center of the sphere, I take a Gaussian sphere and find that the electric field on its surface is E.

Now I enlarge the large sphere, while keeping charge density constant. That Gaussian sphere hasn't changed in size, so E remains constant. As the radius of the large sphere approaches infinity, E doesn't approach 0; it remains constant. The symmetry argument fails because for any finite spherical distribution, "towards the center" is distinguishable from "away from the center", and as the radius of the sphere approaches infinity, the amount of new charge that must be added also approaches infinity.

@ideasrule
I understand that it is physically impossible to have an infinite charge distribution. If you are suggesting that since it is physically impossible, so it is not certain that physical laws will hold good, then please support it with proper reasoning.
I would only like to address the argument you have given for the failure of symmetry, since the other things you have mentioned can be argued with an infinite distribution as well.
The symmetry argument fails because for any finite spherical distribution, "towards the center" is distinguishable from "away from the center", and as the radius of the sphere approaches infinity, the amount of new charge that must be added also approaches infinity.
I do not see as to why the indistinguishability of away and towards the center directions should lead to failure of symmetry, this has only increased the symmetry of the problem and hence making it even more important. I also don't see any reason why if the amount of charge being added tends to infinity, should lead to the failure of symmetry.

I did found the answer in some other discussion forum. I am sharing it below if someone wants to know:
Gauss' law comes from the divergence theorem, which states that the divergence of a vector field integrated over a volume is equal to the vector field itself integrated over the boundary of the volume. However, the divergence theorem is only valid if the vector field is "compactly supported", which means that it's equal to zero outside some finite region. An infinite region of charge is not compactly supported, so the divergence theorem doesn't apply.

This is a somewhat subtle and technical thing which isn't encountered much because basically all charge distributions you'll encounter in physics are finite. If you're interested in the technical details, you might try the Wikipedia article on Stokes' theorem, of which the divergence theorem is a special case:
http://en.wikipedia.org/wiki/Stokes'_theorem

hi!

i came across this thread by discussing a similar problem in the forum. i thougt they convinced me that gauss's law is valid in this problem, but it bugged me the rest of the day, so i continoued to think about it. i would appreciate it if you could take look:

https://www.physicsforums.com/showthread.php?p=3966165#post3966165

i think your explanation should hold for my problem too.

thanks!