Gauss' law for concentric circles

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SUMMARY

This discussion focuses on calculating the electric field and charge distribution for a system of concentric conducting spherical shells. The inner shell has a total charge of -2q, while the outer shell carries a charge of +4q. Using Gauss' law, participants derive the electric field in various regions: inside the inner shell (rd). The charge distribution on the inner and outer surfaces of the shells is also clarified, ensuring the electric field inside the conductors remains zero.

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  • Understanding of Gauss' law and its application in electrostatics
  • Familiarity with electric field concepts and charge distribution
  • Knowledge of spherical coordinates and their relevance in electrostatics
  • Basic proficiency in calculus for deriving electric field equations
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  • Study the derivation of electric fields using Gauss' law in various geometries
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Students of physics, particularly those studying electrostatics, electrical engineers, and anyone interested in understanding electric fields and charge distributions in conductive materials.

  • #61
Doc Al said:
Really? Why then doesn't the problem statement say that? If true, that changes everything! (But I doubt it's true.)
I think it's apparent!It seems to me ;my personal opinion.
 
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  • #62
gracy said:
I think it's apparent!It seems to me ;my personal opinion.
Well, good luck with that. If true, all previous answers are incorrect. o0)
 
  • #63
Doc Al said:
If true
I am no one to decide.Tell me it's correct or not?
 
  • #64
gracy said:
I am no one to decide.Tell me it's correct or not?
Did you draw the diagram, based on your interpretation of the problem? Or was the diagram provided to you?
 
  • #65
Doc Al said:
Or was the diagram provided to you?
provided to me as a solution of the problem
 
  • #66
gracy said:
provided to me as a solution of the problem
I believe that whoever drew the diagram was a little sloppy. The +2q label looks like it's on the blue circle, but it was meant to be on the inner surface of the outer conductor.

Here's the deal: The only place charge resides is on the surfaces of the conductors. In between is empty space: no charges floating around.
 
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  • #67
inner surface of outer conductor falls between the two conductors.So charge residing on it will be automatically between the two conductors.
 
  • #68
gracy said:
inner surface of outer conductor falls between the two conductors.
No, it doesn't. How can the surface of something fall outside that something?

gracy said:
So charge residing on it will be automatically between the two conductors.
No, the charge does not leave the conductor to go floating into space.
 
  • #69
Doc Al said:
No, it doesn't. How can the surface of something fall outside that something?
Not getting.It (this thread)is taking too much time.
 
  • #70
gracy said:
But +2q is in between the two conductive shells.
You are told the +4q is on the outer conducting shell, and in the OP you asked why it was split, putting +2q on the inner circle of radius c. So at that point you understood it is on the inside surface of the outer shell.
 
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  • #71
You have two concentric shells. The pink regions are conductors, the electric field in a conductor is zero.
The empty spaces are voids, there are electric fields inside them, except the central part enclosed by the small shell.
The geometry has spherical symmetry, so is the symmetry of the field.
The charge on the shells can move and redistributed, but no charge can leave any of the shells.
Consider a Gaussian sphere inside the innermost void. There are no charge enclosed, the field is zero.
Consider a concentric sphere inside the first pink region. The electric field must be zero, so that Gaussian sphere does not enclose any charge. There is no charge inside the conductor or on the inner surface of the shell.
Consider a sphere just outside the first shell. How much charge does it enclose? What is the electric field at radius r in the region between the shells?
Consider a sphere inside the outer pink region. It is conductor, the field is zero. How much is the enclosed charge then?
This sphere encloses the small shell and the inner surface of the big shell. How much is the charge that compensates the charge of the inner shell?
Where can this excess charge be distributed? It can not be inside the metal. It can not leave the shell. Where is it?
If there is some charge on the inner surface of the big shell, how much is on the outer surface?
 

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  • shellsgauss.JPG
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  • #72
ehild said:
Consider a sphere just outside the first shell
like this?

like.png
 
  • #73
ehild said:
How much charge does it enclose?
You have not shown any charges.
 
  • #74
  • #75
gracy said:
You have not shown any charges.
Shall I refer my op.
 
  • #76
gracy said:
Shall I refer my op.
Yes, it is the original problem. The inner shell has total charge -2q and the outer shell has charge +4q.
 
  • #77
ehild said:
How much charge does it enclose?
-2q.
 
  • #78
ehild said:
Consider a sphere just outside the first shell. How much charge does it enclose?
gracy said:
-2q.
 
  • #79
ehild said:
What is the electric field at radius r in the region between the shells?
That's where I am stuck.
 
  • #80
gracy said:
That's where I am stuck.
What does Gauss' Law say?
 
  • #81
is my post #78 correct?
 
  • #82
gracy said:
is my post #78 correct?
Yes, the enclosed charge is -2q.
 
  • #83
ehild said:
What is the electric field at radius r in the region between the shells?
you mean at p?
G.png
 
  • #84
Yes, at P or at any other point at the same distance from the centre.
 
  • #85
I am not sure about charge enclosed by this gaussian surface.I know that there is -2q but is +2q also inside it?
 
  • #86
That Gaussian surface encloses the small shell, and some from the void. There is -2q charge on the small shell, and no charge in the empty space. Why should be 2q also inside it??
 
  • #87
ehild said:
Why should be 2q also inside it??
What If I increase r?

ehild said:
What is the electric field at radius r in the region between the shells?
it is still between the shells.

t.png
 
  • #88
Yes. So what is the electric field? I would like to see the formula.
 
  • #89
If i refer picture of post #87 it will include +2q charge also
 
  • #90
gracy said:
If i refer picture of post #87 it will include +2q charge also
NO. Why do you think so?
 

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