Finding potential at the center of metal sphere

  • #51
ehild
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ehild, potential is classically defined as zero at infinity. It's always V/r in my book (Halliday & Resnick).
You can add any arbitrary constant to a potential function: it stays potential for the same problem.
For a point charge, it is convenient to choose the zero of potential at infinity, but that can not be always done: Think of the field near infinite line charge. It is inversely proportional to r, the potential is proportional to ln(r) which is not zero at infinity.

ehild
 
  • #52
rude man
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You can add any arbitrary constant to a potential function: it stays potential for the same problem.
For a point charge, it is convenient to choose the zero of potential at infinity, but that can not be always done: Think of the field near infinite line charge. It is inversely proportional to r, the potential is proportional to ln(r) which is not zero at infinity.

ehild

It's still true for every element of charge dq along the infinite line. It's just when you add (integrate) the dq to + and - infinity that the total line potential becomes infinite.

You have to be very careful when you deal with infinities. We are of course not dealing with them in this problem, and the correct expression for the potential of a finite charge is the integration from +/- ∞ to a point where the potential is to be determined: kq/r.

In the case of the infinite line the calculation of the potential is actually still correct. Potential is work per unit charge, and it takes infinite work to move a unit charge from infinity to the line.
 
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  • #53
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ehild, why would image approach not work at other points in the cavity ? When we consider the case of a point charge outside an isolated metal spherical shell , we have the image charge inside the shell. And then the potential everywhere outside is given by these two charges. We no longer talk about the induced charge on the metallic sphere itself.
So in this particular problem, the point charge is inside the spherical shell with some thickness. As you said, we can divide the problem in two parts. One is the charge itself
and the induced charge on the inner wall. Second is the charge on the outer wall which is distributed uniformly. So we can combine the solutions of these two problems by superposition principle. For the first part of the problem, we can fill entire outer space with the metal. So we can place an image charge somewhere within the metal body.
So the potential everywhere within the inner cavity would be given by the charge and the image charge only. And then we can add the constant potential term due to the second part of the problem.
 
  • #54
rude man
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I take back my last paragraph. I'll leave things by repeating that one has to be careful with infinities. An infinite line is an impossibility so it should not be surprising that calculating its potential as the work from infinity to a point r is mathematically unrealistic.
 
  • #55
ehild
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ehild, why would image approach not work at other points in the cavity ?
I did not say that. I have said that your method would not be easy to calculate the potential in an arbitrary point.

When we consider the case of a point charge outside an isolated metal spherical shell , we have the image charge inside the shell. And then the potential everywhere outside is given by these two charges. We no longer talk about the induced charge on the metallic sphere itself.

In that case the potential everywhere outside is given by three charges: The original one, the image charge and an extra charge of equal magnitude and of opposite sign as the image at the centre of the sphere.

So in this particular problem, the point charge is inside the spherical shell with some thickness. As you said, we can divide the problem in two parts. One is the charge itself
and the induced charge on the inner wall. Second is the charge on the outer wall which is distributed uniformly. So we can combine the solutions of these two problems by superposition principle.

I do not think I said it, but it is very clever. I am getting convinced :smile:

For the first part of the problem, we can fill entire outer space with the metal. So we can place an image charge somewhere within the metal body.
So the potential everywhere within the inner cavity would be given by the charge and the image charge only. And then we can add the constant potential term due to the second part of the problem.

If we really can fill the outer space with a metal (what I said but was not sure if we really can) then the first part is like the problem of a grounded sphere. No need to add that extra charge.

I think, you convinced me about the validity of the solution with the image charge method. :smile:

ehild
 
  • #56
ehild
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I take back my last paragraph. I'll leave things by repeating that one has to be careful with infinities. An infinite line is an impossibility so it should not be surprising that calculating its potential as the work from infinity to a point r is mathematically unrealistic.

The potential of a point charge is zero at infinity. The potential in general need not be zero at infinity.
The potential function we use in calculations must be distinguished from the real potential. Infinite line, infinite plane do not exist, but the potential can be approximated near them with a function that do not tend to zero at infinity.

ehild
 
  • #57
rude man
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The potential of a point charge is zero at infinity. The potential in general need not be zero at infinity.
The potential function we use in calculations must be distinguished from the real potential. Infinite line, infinite plane do not exist, but the potential can be approximated near them with a function that do not tend to zero at infinity.

ehild

Yes, the potential of a point charge is zero at infinity. So is the potential of two charges at infinity. So also for any finite number of charges. I haven't calculated it but I bet the potential of any finite-length, finite charge-density line is also zero at infinity.

Like I said - one must be careful with infinities. For all finite cases, kq/r holds, I believe.
 
  • #58
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As I think about this problem, I think the image approach will work even in situations where
the inner spherical cavity of radius [itex]R_1[/itex] is not centered at the center of the whole sphere. As long as we have a spherical cavity within a metallic sphere, the above image approach will work. So after all the discussion on this problem, we can generalise the situation and we can think of applying this to any spherical cavity within any metallic sphere.
 
  • #59
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rude man or ehild, can you confirm my opinion in post # 58
 
  • #60
ehild
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I have confirmed that the image charge method can be applied for a cavity in post #55.
Find the potential at an arbitrary point when the charge q is very near to the centre of the cavity and see the limit a -->0. You can get the potential easily in case of a central charge. Check if the potentials are equal.

ehild
 
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  • #61
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ehild, in post # 58, I am talking about the inner spherical cavity which is not necessarily concentric with the outer spherical shell. You were replying to different question in post # 55. I just generalised that in post # 58.
 
  • #62
ehild
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I see. How do you get the potential at the outer surface then?

ehild
 
  • #63
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Ok

Irrespective of the shape of the cavity inside the metallic sphere, if we place the charge ,q,
inside the cavity, then equal charge q is distributed uniformly on the outer spherical surface.
The cavity and the outside are different worlds altogether as long as we are talking about
electrostatic case. So potential on the outer surface would just be [itex]\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/itex]
 
  • #64
rude man
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Ok

Irrespective of the shape of the cavity inside the metallic sphere, if we place the charge ,q,
inside the cavity, then equal charge q is distributed uniformly on the outer spherical surface.
The cavity and the outside are different worlds altogether as long as we are talking about
electrostatic case. So potential on the outer surface would just be [itex]\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/itex]

I agree with this of course, issac. It's early in the morning for me :grumpy: but that sounds right.

As for your earlier query: as long as you can wrap a spherical Gaussian surface around the cavity and fully contained in metal, your statement must be true, otherwise the E field could not be zero everywhere about the surface. So it makes no difference what the shape of the metal is - it can be quite irregularly shaped - the charges in the cavity have to be distrtibuted so as to null the net field at every point on the Gaussian shell just beyond R1, just as in the concentric case. So the image would not change from the concentric c ase.
 

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