Gauss' law which formula to use

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SUMMARY

The discussion centers on determining the appropriate formula for calculating the electric field (E) due to surface charge density (a) in different scenarios involving charged sheets. The formula E = a/2ε₀ applies to a single non-conducting sheet, while E = a/ε₀ is used when considering the electric field from two sheets, effectively doubling the field strength. The confusion arises from the choice of Gaussian surfaces, which affects the calculation of the electric field. Understanding the relationship between the Gaussian surface and the resulting electric field is crucial for accurate application of Gauss' law.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Familiarity with electric field concepts and surface charge density
  • Knowledge of Gaussian surfaces and their application in physics
  • Basic principles of electrostatics and conducting materials
NEXT STEPS
  • Study the derivation of Gauss' law and its applications in electrostatics
  • Learn about the behavior of electric fields in conducting materials
  • Explore examples of calculating electric fields using different Gaussian surfaces
  • Investigate the implications of superposition in electric fields from multiple charge distributions
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and Gauss' law applications.

teknonjon
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Homework Statement



how do I know which equation to use for electric field? E=a/2e0 or E=a/e0 when a = surface charge density.

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The Attempt at a Solution

 
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first one is for single surface of a charged sheet ... when it is accompanied by another it changes to the second one (therefore it is twice of it)
 
I looked in my physics book and they use the formula E=a/2e0 when there is only one nonconducting sheet. The picture they show is a cylinder which is the Gaussian surface going through the sheet with both ends outside of the sheet. When they use the formula E=a/e0 half of the Gaussian surface is within the sheet so only one end of the cylinder is sticking ouside the sheet. I don't understand how making the changing the Gaussian surface like this would result in one answer for the electric field being only half the other. Shouldn't the electric field be the same no matter what you use for the Gaussian surface?

thank you for your reply earlier
 
What is the electric field within a conducting material?
 

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