Electric field between conducting and non-conducting sheets

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SUMMARY

The discussion focuses on calculating the electric field between two infinite sheets of charge, where the left sheet is non-conducting with a uniform charge density of 3σ and the right sheet is conducting with a charge density of σ. According to Gauss's Law, the electric field to the left of the non-conducting sheet is E = (3σ)/2ε₀, while the electric field between the sheets is E = (3σ - σ)/ε₀ = (2σ)/ε₀. To the right of the conducting sheet, the electric field is 0 due to the properties of conductors. The clarity of the problem statement and the treatment of charge densities are also discussed.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with electric field concepts and charge density
  • Knowledge of the properties of conducting and non-conducting materials
  • Basic mathematical skills for manipulating equations involving ε₀
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn about induced charge densities on conductors
  • Explore the concept of electric fields in various configurations of charged sheets
  • Investigate the effects of finite dimensions on electric fields from charged plates
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields generated by charged sheets in both theoretical and practical contexts.

Mirza Danish Baig
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Homework Statement



Two infinite sheets of charges are placed parallel to each other. If the sheet on the left is non conducting and have a uniform charge density 3(sigma) and the one on the right is conducting and has a uniform charge density (sigma). If the area on both plates is 1m^2 then calculate the value of electric field at (a) to the left of left sheet (b) in between the sheets (c) to the right of right sheet.

Homework Equations


gauss law E= (sigma)/2€o
E=(sigma)/€o


The Attempt at a Solution


I am attempting it to do it by definitions. I think that the field at the right of the right sheet will be 0 because right sheet is conducting but I don't know if at the left of left sheet the value will be sum of two fields of the sheets or not and what will be the electric field at middle
 
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Mirza Danish Baig said:
the one on the right is conducting and has a uniform charge density (sigma)
This is unclear: Does this include the charge density on both sides of the plate? (I would assume so: The total charge density is sigma.)

Mirza Danish Baig said:
I think that the field at the right of the right sheet will be 0 because right sheet is conducting
The field within the material of the conducting plate must be zero, but not outside it.
 
Doc Al said:
This is unclear: Does this include the charge density on both sides of the plate? (I would assume so: The total charge density is sigma.)
yes I guess so. But its not mention in the question.
 
So you're saying that the charged sheet has a surface charge of 3σ and the conducting sheet has a charge of σ?
 
Mirza Danish Baig said:
Two infinite sheets of charges
Mirza Danish Baig said:
If the area on both plates is 1m^2
??
 
Yeah, quite a bit about this problem statement is not quite clear. Can you post it exactly as it appears, word for word?
 
Doc Al said:
you're
yes. they have given us symbols inspite of numeric values. We have to drive formulas for it. and that's the exact wordings that are written in my note book.
 
Well, ignoring the fact that the "infinite" sheets have an area of 1 m^2, you can just treat them as infinite sheets of charge. Unless there is more to this question, the fact that one is a conducting sheet is irrelevant.
 
I think if the conducting sheet has some small width ##\Delta x## then we ll have to calculate the induced charged density on it and we can speak of the inner and outer surface charge densities on it.. However If its width is zero , so it is purely 2Dimensional then we don't have to bother with that.
 
  • #10
Assuming you're given the total surface charge on the conducting plate (both sides total), you won't have to worry about any induced surface charges since they don't ask for that.
 

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