Gauss' law for concentric circles

In summary, a small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has total charge -2q and the outer shell has charge +4q. Using Gauss' law, the electric field is calculated for various distances r from the common center of the two shells. The distribution of charge on the larger shell is necessary to ensure a zero electric field inside the conductor. The total charge on the inner surface of the small shell is -2q, while the total charge on the outer surface is +2q. On the inner surface of the larger shell, there is +2q charge to cancel out the -2q charge on
  • #36
gracy said:
And sign confusion is still there .Why for r=d it's positive and for r=b &r=c it's negative.What pattern does it follow?from outside to inside or from inside to outside?
Consider the sign of the charge enclosed by Gaussian spheres at those points.
 
Physics news on Phys.org
  • #37
Doc Al said:
Consider the sign of the charge enclosed by Gaussian spheres at those points.
But it also depends on
BvU said:
E2⊥−E1⊥
BvU said:
E1⊥−E2⊥
 
  • #38
gracy said:
But it also depends on
That's an alternate way of looking at it. Mine is easier.
 
  • #39
Doc Al said:
That's an alternate way of looking at it
So which formula are you using?
 
  • #40
gracy said:
So which formula are you using?
I don't need a 'formula' to tell me the direction of the electric field! If the enclosed charge is positive, the field points outward; if negative, inward.
 
  • #41
Doc Al said:
I don't need a 'formula' to tell me the direction of the electric field
E2⊥−E1 or E1⊥−E2
 
  • #42
DocAl's way is actually more practical. But if you insist on knowing how to really decide between above - below or the other way around, I will go into a slightly deeper math, which I hope you can still grasp. See the picture I attached. The gray plane surface is actually a multiple zoom in result of a certain portion of your spherical conducting surface. Make a cylindrical volume such that part of it lies above the conductor, other part lies below it. We have Gauss law ##\oint \mathbf{E}\cdot d\mathbf{a} = Q_{enc}/\epsilon_0 = \int \mathbf{E}_{above}\cdot d\mathbf{a} + \int \mathbf{E}_{below}\cdot d\mathbf{a}##, the integral over the side surface of the cylinder should be zero. To answer your confusion, we actually really have to know the direction of the fields in each region. So, the minus sign in ##E_{2\perp} - E_{1\perp}## can even be a positive sign, depending on the relative directions of the fields in the two regions. The source I was using when I brought up that equation for the first time assumes that the field in both regions point in the same direction, so the minus sign appears. As a conclusion, to determine the correct sign you need to know the direction of the fields in each region. In your problem, the charge on surface r=b is negative, so the fields outside it must point inward.
 

Attachments

  • Gaussian surface.png
    Gaussian surface.png
    4.5 KB · Views: 347
  • Like
Likes gracy
  • #43
blue_leaf77 said:
See the picture I attached
where is it?
 
  • #44
blue_leaf77 said:
DocAl's way is actually more practical.
But I did not understand his way.
 
  • #45
gracy said:
I think I am confusing the point P2 with b.Electric field at r=b was never zero.
gracy said:
But for b &c E is zero.
right?
 
  • #46
blue_leaf77 said:
E1⊥
it does not have any magnitude so no direction.And E2⊥ will have negative sign.
 
  • #47
Doc Al said:
If the enclosed charge is positive, the field points outward; if negative, inward.
if no charges are enclosed?how am I going to decide direction?
 
  • #48
gracy said:
if no charges are enclosed?how am I going to decide direction?
If there is no surface charge density, the normal field components are continuous.
 
  • #49
gracy said:
if no charges are enclosed?how am I going to decide direction?
If no charges are enclosed within your Gaussian sphere (and there is symmetry), then what will the field at the surface be?
 
  • #50
Doc Al said:
then what will the field at the surface be?
Zero.
 
  • #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.
 
  • #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.
 
  • Like
Likes gracy
  • #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.
 
  • Like
Likes gracy
Back
Top