# Applying Guass' Law to Cylindrical Symmetry

1. Feb 1, 2010

### raytrace

OK, having some trouble wrapping my head around this so would appreciate some clarification.

Let us say I had a long, thin wall metal tube of radius R with a uniform charge per unit length. Would there be some magnitude of E of the electric field at a radial distance of R/2?

I understand that there would be no net flux using a gaussian surface smaller than the radius of the tube. I also understand that at the center of the tube, E would be zero. However, I would think that at the radius of R/2 there would be some electric field there.

Could someone explain to me as to why or why not there is E at R/2?

2. Feb 1, 2010

### Staff: Mentor

Since you realize that there's no charge enclosed in a gaussian surface at that radius, why would you think there would be a non-zero E?

3. Feb 1, 2010

### raytrace

Well, a net zero flux from a gaussian surface just means that you have an equal number of electric field lines going in as going out. Just because the flux is a net zero doesn't mean that there doesn't exist an electric field, right?

4. Feb 2, 2010

### ThomasYoung

The flux is a net zero does mean that there doesn't exist an electric field!

5. Feb 2, 2010

### Staff: Mentor

Right, but the symmetry of the charge distribution allows you to make a much stronger statement. In this case, due to the cylindrical symmetry of the charge distribution, you know that at any radius the field must be radial and uniform. It's that symmetry combined with Gauss's law that allows you to conclude that the field is zero for r < R.

In general you are correct. But in this case it does mean that there is no electric field for r < R.