Electric field inside a charged cylinder

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SUMMARY

The electric field inside a hollow charged cylinder with charge Q on its outer surface is zero at any point along the center axis when applying Gauss' law, as there is no enclosed charge. However, this does not imply the absence of an electric field; rather, the contributions to the electric field from differential rings of charge cancel each other out. The correct approach to determine the electric field at a point A on the center axis involves summing the contributions from these differential rings, rather than relying solely on Gauss' law. The electric field can be calculated using the formula E = k*Q*r-hat/r^2, acknowledging that the field varies along the length of the cylinder.

PREREQUISITES
  • Understanding of Gauss' law
  • Familiarity with electric fields and their calculations
  • Knowledge of differential calculus for summing contributions
  • Basic concepts of charge distribution in cylindrical geometries
NEXT STEPS
  • Study the application of Gauss' law in non-uniform electric fields
  • Learn how to calculate electric fields from continuous charge distributions
  • Explore the concept of electric field lines and their behavior in cylindrical coordinates
  • Investigate the effects of different charge distributions on electric fields
USEFUL FOR

Students and educators in physics, particularly those focused on electromagnetism, as well as anyone seeking to deepen their understanding of electric fields in charged geometries.

yoni162
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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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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?
 
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
 
yoni162 said:
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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