Gauss' law for concentric circles

AI Thread Summary
The discussion revolves around applying Gauss' law to analyze the electric field and charge distribution in a system of concentric conducting spherical shells. The inner shell has a charge of -2q, while the outer shell has a total charge of +4q, which is distributed as +2q on its inner surface and +2q on its outer surface to maintain zero electric field within the conductor. The electric field is calculated for various regions: it is zero inside the inner shell, has a specific magnitude between the shells, and is directed outward in the region outside the outer shell. The participants clarify the differences in electric field behavior at points P3 and P4, emphasizing that the field is non-zero at P3 due to enclosed charges, while it is zero at P4 because it lies on the conductor's surface. Overall, the conversation highlights the importance of understanding charge distribution and electric field direction in electrostatics.
  • #51
gracy said:
But I did not understand his way.
It's easy, if you use Gauss' law.

For example, say you want the field in the space between the two conducting shells. So draw a Gaussian sphere in that region. What is the charge contained within that Gaussian surface? Answer that and we can proceed.
 
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  • #52
gracy said:
Zero.
Right!
 
  • #53
Doc Al said:
What is the charge contained within that Gaussian surface?
The charge enclosed by one of the(inner) conductors.
 
  • #54
gracy said:
The charge enclosed by one of the(inner) conductors.
Give me the charge in terms of q.
 
  • #55
Doc Al said:
Give me the charge in terms of q.
Shall I refer picture in op.
 
  • #56
gracy said:
Shall I refer picture in op.
Yes, that's the problem we are discussing. The one you started the thread with.
 
  • #57
Zero.Because there is +2q and -2q.
 
  • #58
gracy said:
Zero.Because there is +2q and -2q.
No.

I suspect the diagram is throwing you off. For some reason, there is a blue circle drawn in the space between the two conducting shells (where p3 is). That circle is labeled as "+2q", implying that there is some charge just floating in space. I doubt that's what you meant. I think you meant to show that the inner surface of the outer shell has a charge of +2q.

Until this is straightened out there is little point in continuing.
 
  • #59
But +2q is in between the two conductive shells.
Doc Al said:
For example, say you want the field in the space between the two conducting shells.
 
  • #60
gracy said:
But +2q is in between the two conductive shells.
Really? Why then doesn't the problem statement say that? If true, that changes everything! (But I doubt it's true.)
 
  • #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.
 
  • #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?
 

Attachments

  • shellsgauss.JPG
    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?
 
  • #91
ehild said:
NO. Why do you think so?

y.png
 
  • #92
gracy said:
View attachment 89159
Once again, that +2q charge is on the surface of the outer conductor, not in the space between conductors.

From r > b to r < c there is no charge, only empty space.
 
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  • #93
Doc Al said:
that +2q charge is on the surface of the outer conductor, not in the space between conductors.
Ok.I finally understand.Can we proceed?
 
  • #94
ehild said:
What is the electric field at radius r in the region between the shells?
It would be - 2q/4πε0r^2
 
  • #95
ehild said:
Consider a sphere inside the outer pink region. It is conductor, the field is zero. How much is the enclosed charge then?
zero.
 
  • #96
ehild said:
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?
+2q
 
  • #97
ehild said:
It can not leave the shell. Where is it?
on the outer surface.
 
  • #98
ehild said:
If there is some charge on the inner surface of the big shell, how much is on the outer surface?
+2q.Now I understood the complete setup of the problem.
 
  • #99
I am happy :oldsmile:
 
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  • #100
Wait.
what is the distance of -2q (distributed throughout the outer surface of inner conductor) from the center of inner shell?How can it be b?
 

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