Electric flux trough an infinite plane 2

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SUMMARY

The discussion revolves around calculating the electric flux through an infinite plane due to a point charge Q located at the origin. The primary focus is on applying Gauss's Law, which simplifies the process by relating electric field lines to flux without the need for complex integration. A key point raised is the importance of correctly using the differential area element dS, specifically noting that dS equals 2πr dr. The participants emphasize clarity in presenting solutions, suggesting the use of TeX for better readability.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric field concepts
  • Knowledge of differential area elements in calculus
  • Ability to interpret and create mathematical representations using TeX
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Learn about electric field calculations for point charges
  • Explore the use of TeX for mathematical documentation
  • Review differential area elements in three-dimensional integrals
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Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking to enhance their teaching materials with clear mathematical representations.

julius71989
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Sorry, what I posted earlier was wrong.. here is the question:

an charge Q is at the origin. Find the electric flux an infinite plane at z distance from z axis.

Please check my attempts to the solution. Please check what's wrong in it because until now I am not getting the correct answer.. thanks a lot. click the link below:

figure: http://pic40.picturetrail.com/VOL359/11337670/20318718/330158477.jpg
solution: http://pic40.picturetrail.com/VOL359/11337670/20318718/330158480.jpg
 
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Man! That is REALLY hard to read. I usually don't pester people about using TeX, but that's pretty bad. The sheer font wackiness is awful. I think you are forgetting that dS=2*pi*r*dr. Note the r. I'd be more sure, but I can't look at that jpg again. It hurts my eyes. :)
 
Last edited:
You don't have to integrate it to find the flux. Think in terms of Gauss law and electric field lines and you can avoid all the maths.
 

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