Electric field inside a charged cylinder

  • Thread starter yoni162
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  • #1
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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?
 

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Answers and Replies

  • #2
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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?
 
  • #3
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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
 
  • #4
Doc Al
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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.
 

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