Electric field between 2 plates

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SUMMARY

The electric field between two charged plates is defined as σ/(2ε₀) for each plate, where σ represents the surface charge density and ε₀ is the permittivity of free space. When both plates possess equal and opposite charges, the total electric field is the sum of the contributions from each plate, resulting in E = σ/ε₀. This derivation is grounded in Gauss's Law, which is essential for understanding the behavior of electric fields in this configuration.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and surface charge density
  • Basic knowledge of electrostatics concepts
  • Awareness of permittivity of free space (ε₀)
NEXT STEPS
  • Study the derivation of electric fields using Gauss's Law
  • Explore the concept of electric flux in electrostatics
  • Learn about the applications of Coulomb's Law in electric field calculations
  • Investigate the effects of different charge configurations on electric fields
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators and anyone seeking to deepen their understanding of electric fields and their derivations.

staetualex
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Why electric field between 2 charged plates is sigma(surface charge density)/(2epsilonO) permitivity of free space?
 
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You are wrong brother..
The electric field between 2 charged plated is sigma/epsilonO.. provided both the plates have equal and opposite charge..
This is proved using Gauss's Law.
If u haven't learned about this concept.. search Wikipedia or any other good book.

Cheers
 
Yeah, sorry, one plate comes with sigma/2epsilon0, another comes with sigma/2epsilon0, add them together and we get sigma/epsilon0. Anyone wants to share why is the electric field sigma/2epsilon0?
 
staetualex said:
Yeah, sorry, one plate comes with sigma/2epsilon0, another comes with sigma/2epsilon0, add them together and we get sigma/epsilon0. Anyone wants to share why is the electric field sigma/2epsilon0?

Your question is vague. Are you asking HOW one derives such an electric field for that configuration? If you are, then you already had your answer via Gauss's law. If you are asking on how to make such a derivation, then you need to make further elaboration on your academic background, i.e. have you been taught Gauss's law and how to apply it? And this needs to be done in the HW/Coursework forum.

Zz.
 
Finished high school, no college (19). I was studying Coloumb's law, electric flux and so on, and i stumbled upon that formula. Just wanted to know why the formula valid. That link, with gauss law, it may help me prove myself, just need to crunch the numbers.
 

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