- #61
gracy
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I think it's apparent!It seems to me ;my personal opinion.Doc Al said:
Gauss' law for concentric circles
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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.
- #62
Doc Al
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Well, good luck with that. If true, all previous answers are incorrect.gracy said:
- #63
gracy
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I am no one to decide.Tell me it's correct or not?Doc Al said:
- #64
Doc Al
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Did you draw the diagram, based on your interpretation of the problem? Or was the diagram provided to you?gracy said:
- #65
gracy
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provided to me as a solution of the problemDoc Al said:
- #66
Doc Al
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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.gracy said:
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.
- #67
gracy
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inner surface of outer conductor falls between the two conductors.So charge residing on it will be automatically between the two conductors.
- #68
Doc Al
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No, it doesn't. How can the surface of something fall outside that something?gracy said:
No, the charge does not leave the conductor to go floating into space.gracy said:
- #69
gracy
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Not getting.It (this thread)is taking too much time.Doc Al said:
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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.gracy said:
- #71
ehild
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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?
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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- #72
gracy
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like this?ehild said:
- #73
gracy
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You have not shown any charges.ehild said:
- #74
ehild
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yes.gracy said:
- #75
gracy
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Shall I refer my op.gracy said:
- #76
ehild
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Yes, it is the original problem. The inner shell has total charge -2q and the outer shell has charge +4q.gracy said:
- #77
gracy
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-2q.ehild said:
- #78
gracy
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ehild said:
gracy said:
- #79
gracy
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That's where I am stuck.ehild said:
- #80
ehild
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What does Gauss' Law say?gracy said:
- #81
gracy
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is my post #78 correct?
- #82
ehild
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Yes, the enclosed charge is -2q.gracy said:
- #83
gracy
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you mean at p?ehild said:
- #84
ehild
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Yes, at P or at any other point at the same distance from the centre.
- #85
gracy
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I am not sure about charge enclosed by this gaussian surface.I know that there is -2q but is +2q also inside it?
- #86
ehild
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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
gracy
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What If I increase r?ehild said:
it is still between the shells.ehild said:
- #88
ehild
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Yes. So what is the electric field? I would like to see the formula.
- #89
gracy
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If i refer picture of post #87 it will include +2q charge also
- #90
ehild
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NO. Why do you think so?gracy said:
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