# Gauss' law for uniformly charged space

Likith D
the problem:
Say we have the entire space uniformly charged. Then, the E field experienced by any point is zero, from symmetry.*
But, it means that for any Gaussian surface, the flux though it is zero even though the charge enclosed is clearly not. Gauss' law seems to disagree with symmetry, but it also cannot 'therefore state' that such a charge distribution is not possible.** (only theoretically, if it should exist)

*It is not 'not defined' for the same reasons E field inside a sphere of uniform charge distribution is not 'not defined'... so to say that it is not zero is to go against symmetry of space for that point
**Why would we not have 3d infinite charge distribution while we have 2d infinite charge distribution

attempt at solution:
So, I tried to use the fact that we already computed E field inside a spherical uniformly charged object and let the R tend to infinity which gives ; https://i.stack.imgur.com/N8dwe.jpg independent of radius of sphere.
which may or maynot be zero depending on center of sphere and the point, which makes it weirder... but I have tried integration to find the E field of a uniformly charged wire segment and made it's length tend to infinity to get an answer that agrees with Gauss' law (the same for a planar disc tending to infinite plane, works)... and uniformly charged space seems to be not following that...
Gauss' law just seems to disagree with uniform charged space
what do we make of all this? that Gauss' law is flawed?
If it cannot possibly go against symmetry, does it really imply that uniformly charged space is not possible?

## Answers and Replies

Staff Emeritus
Homework Helper
Gold Member
You need boundary conditions, i.e., a behaviour at infinity, that breaks the symmetry in order for Gauss' law to be consistent. This was discussed relatively recently in a featured thread. Boundary conditions that satisfy either translational or rotational invariance will break the other.

Last edited: