Charge density on the surface of a conductor

  • #1
marcos7615
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Homework Statement:
There is a flat conductor, charge Q, very large surface area S and thickness
“a“ parallel to a surface charge distribution of density σ = 2Q/S
and separated a distance "d" from it, as shown in the figure. It is requested:

- The charge density σ1 on the outer surface of the conductor
of the load distribution.
Relevant Equations:
σ = 2Q/S
1653165597565.png


I have tried to solve the problem by setting as a condition that the electric field inside the conductor has to be 0, but in this way I have two unknowns (σ1 and σ2):
1653166686126.png
 

Answers and Replies

  • #2
TSny
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I have tried to solve the problem by setting as a condition that the electric field inside the conductor has to be 0, but in this way I have two unknowns (σ1 and σ2):
1653168712217.png

Check the signs of the terms on the left side.

Also, you should be able to come up with another relationship between ##\sigma_1## and ##\sigma_2## using the fact that the total charge on the conductor is given to be ##Q##.
 
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  • #3
alan123hk
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Will it look like the picture below?
A12.jpg
 
  • #4
TSny
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Will it look like the picture below?
I don't understand this picture. Volume charge density ##\rho## is not relevant to this problem.

You made a good start in your first post where you got a relation between the surface charge densities based on the fact that ##E## must equal zero inside the conductor. However, there is a sign error in your relation
1653484077758.png


Once you make the correction, this relation gives you one equation for the two unknowns ##\sigma_1## and ##\sigma_2##. The total charge ##Q## of the conductor and the the charge density ##\sigma## of the plane are considered as known.

Try finding another equation involving ##\sigma_1## and ##\sigma_2## based on the idea that the net charge of the conductor is ##Q##.
 
  • #5
alan123hk
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Sorry I was in a hurry so the description in my previous post was wrong. By the way, I'm not the OP of this thread, I'm just interested in this question.

This is my answer after much deliberation. I believe this time should not go wrong.

A13.jpg
 
  • #6
TSny
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By the way, I'm not the OP of this thread, I'm just interested in this question.
Yes, I mistakenly assumed you were the OP. It has been several days without any response from the OP.

Your solution is correct.
 
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