Electric field inside a charged cylinder

[yoni162]

Homework Statement

A cylinder (hollow) with radius R is charged with charge Q on its outer side. What is the electric field at a point A on its center axis (inside the cylinder)?

Homework Equations

Electric field generated by a charged ring in distance r from its center
Gauss' law

The Attempt at a Solution

I seem to have misunderstood something here..I could look at a ring of width dz and calculate its contribution to the electric field at the point A on the center axis of the cylinder. This will definitely sum up to something, meaning there will be a field at point A.
On the other hand, if I use Gauss' law, I take a cylinder surface of radius r<R surrounding the center axis. Obviously Qin=0, so by conclusion the electric field in any point where r<R is 0. where's my mistake?

 

[yoni162]

Alright I think I've realized my misconception..Gauss' law in this case only tells my the the electric flux will be 0, since there is no charge inside the surface I chose..this doesn't mean that there isn't an electric field there, it just means that all contributions of flux cancel each other out, no? The electric field is there, but it cannot be calculated with Gauss' law, only directly by summing all the field contributions due to the differential rings which makw up the cylinder. Am I correct?

 

[Robin Lee]

You will definitely get 0 if you use Gauss's Law due to the flux being zero. I think you can use E=k*Q*r-hat/r^2

 

[Doc Al]

Alright I think I've realized my misconception..Gauss' law in this case only tells my the the electric flux will be 0, since there is no charge inside the surface I chose..this doesn't mean that there isn't an electric field there, it just means that all contributions of flux cancel each other out, no? The electric field is there, but it cannot be calculated with Gauss' law, only directly by summing all the field contributions due to the differential rings which makw up the cylinder. Am I correct?

Yes, your thinking is correct. Gauss' law always applies, but it's not always helpful. It's helpful when symmetry tells you that the field along the surface is uniform, but that's not the case here. Here the field is different near the ends of the cylinder compared to the middle.

 

[rwisz]

Your approach using Gauss' law is correct. The mistake in your thinking is that the electric field inside the charged cylinder is not zero. This is because, while there is no net charge inside the cylinder, there is still a non-zero electric field at every point inside the cylinder due to the charge on the outer surface. This electric field is directed radially towards the center of the cylinder and its magnitude can be calculated using Gauss' law. The fact that the net charge inside the cylinder is zero does not mean that there is no electric field inside the cylinder.