Comparing Flux of a Closed Cylinder

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SUMMARY

The discussion focuses on calculating the electric flux through the ends of a closed cylinder when a negatively charged particle is positioned outside its axis. The flux through the left end of the cylinder is determined to be positive, while the flux through the right end is negative due to the direction of the electric field. The participants emphasize the importance of integrating the flux through differential annuli at both ends of the cylinder and comparing the results to understand the effects of distance on electric flux.

PREREQUISITES
  • Understanding of Gauss's Law for electric flux
  • Familiarity with cylindrical coordinates and integration techniques
  • Knowledge of electric fields generated by point charges
  • Basic concepts of electric charge and its effects on surrounding space
NEXT STEPS
  • Study Gauss's Law and its applications in electrostatics
  • Learn about electric field calculations for point charges
  • Explore integration techniques in cylindrical coordinates
  • Investigate the effects of distance on electric flux in various geometries
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone seeking to deepen their understanding of electric flux and its calculations in closed geometries.

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Homework Statement


A closed cylinder consists of two circular end caps and a curved side. A negatively charged particle is placed outside the left end of the cylinder, on its axis. Compare the flux through the left end of the cylinder with that through the right.


2. The attempt at a solution

I know the field points out of the cylinder on the left end, and into the cylinder on the right, so the flux through the left end is positive while that through the right end is negative. But what above the sides of the curves?
 
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Does the question ask you about the curved sides?
What law tells you how the electric flux from a point charge, through a fixed area, changes with distance?
 
Let
charge located at x = 0
near end locared at x = d
length of cylinder = L
R = radius of ends

1. Compute the flux through a differential annulus for the near end. Integrate from r = 0 to r = R parametrically using θ = arc tan r/d.
2. Do the same for the far end. Realize θ will range from 0 to a smaller number than for the near end.
3. Compare.
 
Thanks !
 

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