Gauss' law which formula to use

Click For Summary

Homework Help Overview

The discussion revolves around the application of Gauss' law in determining the electric field due to surface charge density. Participants are exploring the appropriate formulas to use in different scenarios involving charged sheets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which different formulas for electric field are applicable, particularly in relation to single versus multiple charged sheets. Questions arise about the implications of using different Gaussian surfaces and how that affects the calculated electric field.

Discussion Status

The conversation is ongoing, with participants providing insights into the use of specific equations based on the configuration of charged sheets. There is an exploration of how the choice of Gaussian surface influences the results, but no consensus has been reached regarding the interpretation of these differences.

Contextual Notes

Some participants reference specific scenarios from textbooks, indicating a reliance on visual aids and examples to understand the concepts. There is also a mention of confusion regarding the behavior of electric fields within conducting materials, suggesting a need for further clarification on that topic.

teknonjon
Messages
2
Reaction score
0

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.

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
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?
 

Similar threads

Gauss' Law: Understand How to Calculate Flux
  • · Replies 1 ·
Replies
1
Views
2K
Use of imaginary charge vs Gauss' law
  • · Replies 12 ·
Replies
12
Views
1K
Electric field due to charges between 2 parallel infinite planes
  • · Replies 9 ·
Replies
9
Views
2K
I Thought I Understood Gauss' Law (Sheets)
  • · Replies 26 ·
Replies
26
Views
3K
Gauss' law problem -- Should the field due to both the inside and outside charges be taken into account?
  • · Replies 28 ·
Replies
28
Views
2K
No Electric Charges if No Electric Field in Region
  • · Replies 22 ·
Replies
22
Views
3K
Flux of Electric field through sphere
  • · Replies 2 ·
Replies
2
Views
2K
Gauss' Law for a conducting / non-conducting sheet
  • · Replies 5 ·
Replies
5
Views
2K
Applying Gauss' Law for Calculating Work Req'd to Move Charge
  • · Replies 9 ·
Replies
9
Views
2K
Find the density of surface charges at the boundary of two conductors with different resistivities
  • · Replies 6 ·
Replies
6
Views
2K