Finding potential at the center of metal sphere

  • #1
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Hi

Here is a problem I am trying to do. A point charge q is located at a distance r
from the center O of an uncharged conducting spherical layer whose inside and
outside radii are equal to [itex]R_1[/itex] and [itex]R_2[/itex] respectively.
Find the potential at the point O if [itex]R_1 < R_2[/itex].

Now I was thinking of method of images. But since we have a spherical conducting
sphere with some thickness, that would mean we have several concentric equipotential
surfaces surrounding the chrge q. So one image charge would not suffice. We will
need infinitely many of the image charges. So method of images is not practical here. Griffiths
says in his book that the leftover charge on the outer surface in case of a charge
placed in a metal cavity is uniformly distributed. He doesn't give any satisfactory
reasoning. But lets assume what he says. Then the outside of the metal sphere, world
will see that charge q is at the center of the sphere, which means potential outside
is given by

[tex]V(r)=\frac{1}{4\pi\epsilon_o}\frac{q}{r}[/tex]

where r is the distance of any point outside the sphere from the center of
the sphere. So on the outer wall of the sphere, the potential is

[tex]V(R_2)=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/tex]

since sphere is metal , its equipotential, so the potential on the inner wall of the
sphere is same. So what else can we say so that we can get the potential at the
center of the sphere ?

thanks
 

Answers and Replies

  • #2
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you can do it by the legendre polynomials or method of images. If you're having problems still i think there is a good example of it in griffith's introduction of e&m textbook
 
  • #3
912
19
Hi

I think I got it. We don't need advanced techniques like differential equations. Charge -q will
be induced on the inner wall of the sphere and so, charge +q will be left over the outer wall.
Now we can just sum over the potential contributions from three. For example , for
the inner wall, the potential at the center will be given by

[tex]V_{inner}=\frac{1}{4\pi\epsilon_o}\int \frac{dq}{r}[/tex]

[tex]V_{inner}=\frac{1}{4\pi\epsilon_o}\int \frac{\sigma(\theta,\phi)\;dq}{r}[/tex]

Here , for each surface element, r from the center remains constant, [itex]R_1[/itex]. So
remaining integral just integrates to the total induced charge, which is -q.

[tex]V_{inner}=\frac{1}{4\pi\epsilon_o}\frac{-q}{R_1} [/tex]

Similarly for the charge on outer wall, which is q, the potential at the center would be

[tex]V_{outer}=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2} [/tex]

so the total potential at the center would be (adding the potential contribution due to
the charge q itself)

[tex]V=\frac{1}{4\pi\epsilon_o}\left[\frac{q}{r}+\frac{q}{R_2}-\frac{q}{R_1}\right] [/tex]

I took some efforts to get this....
thanks
 
  • #4
rude man
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I wonder about this. My problem is that if we let R1 = 0 in your result we get infinite potential. I looked up this situation1) for a solid sphere which of course corresponds to R1 = 0. The image for this has magnitude -qR2/r and located a distance (R2)2/r from O towards r.

The potential for this configuration is readily calculated to be -kq/R2 + kq/r which certainly isn't infinite. Here k = 1/4πε0.

Now, my contention is that R1 does not enter the computation. That's because if I take a unit test charge along a line from -∞ toward O (with q beyond O), once I hit the outer surface R2 the potential doesn't change any further until I hit O.

Bottom line, I think the answer is V = -kq/R2 + kq/r.

1) Ramo & Whinnery, Fields and Waves in Modern Radio, Wiley, 2nd ed. p. p.51
 
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  • #5
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I wonder about this. My problem is that if we let R1 = 0 in your result we get infinite potential. I looked up this situation1) for a solid sphere which of course corresponds to R1 = 0. The image for this has magnitude -qR2/r and located a distance (R2)2/r from O towards r.

The potential for this configuration is readily calculated to be -kq/R2 + kq/r which certainly isn't infinite. Here k = 1/4πε0.

Now, my contention is that R1 does not enter the computation. That's because if I take a unit test charge along a line from -∞ toward O (with q beyond O), once I hit the outer surface R2 the potential doesn't change any further until I hit O.

Bottom line, I think the answer is V = -kq/R2 + kq/r.

1) Ramo & Whinnery, Fields and Waves in Modern Radio, Wiley, 2nd ed. p. p.51

that is why i suggested the "legendre" method because that allows for the R=0 case
 
  • #6
912
19
If we consider the limit of [itex]R_1[/itex] going to zero, then we will need another
way of doing this. For method of images, where will you put this charge q, because
all the free charge goes to the surface of the conductor.
 
  • #7
rude man
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An "image" charge is a ficticious entity. The idea is that the image replaces the conductor . In other words, the image is part of a model representing the conductor (the rest of the model in your case is of course q). An image is not placed into the conductor. You ditch the conductor and replace it with the image charge so you can then employ conventional point-charge analysis.
.
 
  • #8
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But, when you reduce [itex]R_1[/itex], you have to reduce r as well, otherwise you are
changing the conditions of the problem. So, when both [itex]R_1[/itex] and r go to zero,
[itex]R_1 \approx r[/itex] , so the first two terms cancel and you do get the potential
of the charge q on a metal sphere.
 
  • #9
rude man
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But, when you reduce [itex]R_1[/itex], you have to reduce r as well, otherwise you are
changing the conditions of the problem. So, when both [itex]R_1[/itex] and r go to zero,
[itex]R_1 \approx r[/itex] , so the first two terms cancel and you do get the potential
of the charge q on a metal sphere.

OK, I suppose q could be located inside R1. Is that the case? If so, the potential at O would be kq/R2 + kq/r - kq/(R1 + r).
 
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  • #10
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I just rechecked the problem. I didn't write the last line of the problem correctly.
It should read .... Find the potential at the point O if [itex]r < R_1[/itex]. I had this in mind
when I solved the problem, but its only now that I realise that I made a mistake
while typing the problem..
 
  • #11
rude man
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OK. I should have considered that possibility too. I thought r > R2.

Now, I wonder if our answers are really the same? I'll look into it.

And I agree, images are of no help in that case.
 
  • #12
rude man
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No, they're not the same. You have the term kq/R1
wheras I have the term kq/(R1 + r).
 
  • #13
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sorry for the confusion.

By the way I have related question about this problem. If we have a cavity(of some weird shape) in spherical metal ball and if we place charge in it, then we of course have induced charge on the cavity walls and then we have equal but opposite charge
on the outer surface on the sphere. Now Griffiths in his books says that the charge
on the outer surface get distributed uniformly irrespective of the cavity shape and the
location of test charge inside the cavity. He didn't really offer any explanation for this.
What would be the reason ?
 
  • #14
rude man
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The basic thing is that charges of the same polarity will always try to get as far away from each other as possible. If the charge distribution were not uniform then there would be areas where the charges are closer together than they have to be, and they would redistribute until the charge density is uniform per unit area.

I read your argument on how you reached your answer. Didn't necessarily understand it. My approach was different. I placed q on the x-axis to the right of O, 0<r<+R1, then asked what work had to be done to move a unit positive test charge from minus infinity along the x axis until it sat at O.
 
  • #15
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I think there is some problem with your argument. If the charge is inside the sphere, you will need to cross the walls of the sphere to come from infinity at that point.
 
  • #16
rude man
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I think there is some problem with your argument. If the charge is inside the sphere, you will need to cross the walls of the sphere to come from infinity at that point.

That's right. Where the electric field, so the force and work necessary to do that, and the change in potential, are all zero!
 
  • #17
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(Excuse me if my English is less than perfect. It is not my main language.)

I believe that the potential at the center is rather [tex] \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R_1} \right). [/tex] This is rude man's first answer, except that I have [itex]R_1[/itex] where he has [itex]R_2.[/itex] My reasoning is as follows: you can write the potential inside the sphere as [itex]U_q + U[/itex], where the first term is the potential of the point charge, which is of course known, and the second is the potential of the charge distribution that is induced in the conducting layer. An important difference between these two is that whereas [itex]U_q[/itex] is singular where the point charge is located, [itex]U[/itex] is a harmonic function on the whole inside of the sphere. Moreover, since the full potential needs to be constant on the inner surface, say zero, we have [itex]U = -U_q[/itex] on the inner surface. So now you have an ordinary Dirichlet boundary value problem for [itex]U[/itex]. You could solve this using separation of variables and all that, as suggested by Liquidxlax, but if you are merely interested in the potential at the center then there is a easier way. The potential at the center is [itex]U_q(0) + U(0)[/itex], and the first term is again no problem. For the second term, you can use Gauss' mean value theorem, which says that the value of a harmonic function at the center of a sphere is equal to the average value of the function on the surface of the sphere. But as we just saw, [itex]U = -U_q[/itex] on the inner surface, so the average of [itex]U[/itex] is minus the average of [itex]U_q[/itex]. To do the calculation it is best to use spherical coordinates, with the point charge on the zenith axis. Then [tex]U_q(R_1, \theta, \phi) = \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}}[/tex] (I used the cosine rule for this), and so the average value is [tex] \frac{1}{4 \pi R_1^2} \int_{\textrm{sphere}} U_q \, dS = \frac{q}{16 \pi^2 \epsilon_0} \int_{\theta = 0}^\pi \int_{\phi = 0}^{2 \pi} \frac{\sin \theta}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}} \, d\theta \, d\phi [/tex] [tex] = \frac{q}{8 \pi \epsilon_0} \int_0^\pi \frac{\sin \theta}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}} \, d\theta = \frac{q}{16 \pi \epsilon_0 r R_1} \int_{(R-r)^2}^{(R+r)^2} \frac{du}{\sqrt{u}} = \frac{q}{4 \pi \epsilon_0 R_1}.[/tex] Hence, the potential at the center is [tex]\frac{q}{4 \pi \epsilon_0 r} - \frac{q}{4 \pi \epsilon_0 R_1} = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R_1} \right).[/tex]

I am by no means an expert on electrostatics, so this may all be wrong. I would very much like to hear your opinion on this attempted solution.

With kind regards,
jch1
 
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  • #18
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jch, I didn't understand your solution, but it seems wrong. If we shrink the inner spherical
layer to zero, then as I have pointed out in some last post, both [itex]r[/itex] and
[itex]R_1[/itex] shrink to zero, [itex]R_1\approx r [/itex] and then your solution tends to zero. But what should happen is that potential should tend to

[tex]\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/tex]

because, there is charge q distributed on the outer surface. And my solution achieves that. Your doesn't .
 
  • #19
rude man
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I wish I could follow jch1's argument but I could not, at least for the moment.

I also have a bit of unease about my own derivation:

1: can one really assume that the presence vs. absence of the shield doesn't affect the potential at x = -R2? I think so, but ...

2: this derivation assumes the work function delivers positive kinetic energy (work function) to the test charge as it enters the shell at -R2, and then loses it when it exits at -R1. Seems OK from a conservation-of-energy argument.

So for the moment I'll stick with V(O) = kq{1/r + 1/R2 - 1/(r + R1)}.

Isaac, are you going to find out what the answer is from your instructor? We'd certainly like to know, including the derivation.
 
  • #20
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Is there anything wrong with my solution ? I don't see any logical gaps. Just used superposition principle.

I am not student.... just doing this for fun...
 
  • #21
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Issacnewton: on second thought I think that my solution is indeed wrong and that yours is correct. I assumed that I could take the potential to be zero on the conductor, but your problem statement said nothing about the conductor being grounded. The potential has to be zero at infinity. If we take Griffiths' statement about the uniform distribution of the induced charges on the outer surface for granted, then the potential in the conductor is indeed [itex] \frac{q}{4 \pi \epsilon_0 R_2} [/itex], as you claimed. If I rerun the last part of my argument with this new value, I get: [itex]U(0)[/itex] = average of [itex]U[/itex] on the inner surface = average of [itex]\frac{q}{4 \pi \epsilon_0 R_2} - U_q[/itex] on the inner surface = average of [itex]\frac{q}{4 \pi \epsilon_0 R_2} [/itex] on the inner surface - average of [itex]U_q[/itex] on the inner surface = [itex]\frac{q}{4 \pi \epsilon_0 R_2} - \frac{q}{4 \pi \epsilon_0 R_1}. [/itex] (The second term is the value that I calculated in my first post and on which my error has no bearing.) So the total potential at the center = [tex] U_q(0) + U(0) = \frac{q}{4 \pi \epsilon_0 r} + \frac{q}{4 \pi \epsilon_0 R_2} - \frac{q}{4 \pi \epsilon_0 R_1},[/tex] which is the answer you gave. Moreover, your argument is simpler en much more insightful.

This being said, I do not agree with your argument about shrinking [itex]R_1[/itex]. My erroneous solution does not necessarily tend to zero when [itex]R_1[/itex], and therefore also [itex]r[/itex], do. If you take [itex]R_1 = r + a r^2[/itex] for example, where [itex]a[/itex] is a positive real number, then [itex]\frac{1}{r} - \frac{1}{R_1}[/itex] tends to [itex]a[/itex], if I calculated it right. It seems to me that this limit situation is ill defined: the point charge and the induced charges all get piled on top of one another. But it doesn't matter much, since my solution was wrong anyway.

rude man: it seems to me that the problem with your approach is that to calculate the work done by the field on the charge on the segment inside the sphere (that is, from [itex]R_1[/itex] to [itex]O[/itex]), you need to know the field there, but the field (or the potential) is what we are looking for in the first place. Or do I misunderstand your argument?

With kind regards,
jch1
 
  • #22
rude man
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rude man: it seems to me that the problem with your approach is that to calculate the work done by the field on the charge on the segment inside the sphere (that is, from [itex]R_1[/itex] to [itex]O[/itex]), you need to know the field there, but the field (or the potential) is what we are looking for in the first place. Or do I misunderstand your argument?

With kind regards,
jch1

jch1, your point is well taken and is reflected in my caveat about the shell not affecting the interior field distribution.

My thought: suppose that q is located at x = +r. Then there will be positive charges induced along the inside surface at x = -R1 and negative charges induced along the inside surface at x = +R1. But sincxe q is located closer to x=R1 tan x = -R1, the charge density along x = -R1 will be less than that along x = +R1. Combined with the charge q at location x = +r, the effect of the shield is nullified and the field inside the shell is the same as if the shell did not exist.

I readily admit this is pretty much nothing more than conjecture. I find myself having to admit that I can't solve the problem rigorously. Thank you both, jch1 anfd IssacN, for your insights and help. Sorry I could not contribute more.

Issac, that problem must have come from somewhere. Do you know the right answer? If so it must be yours! :smile:
 
  • #23
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This being said, I do not agree with your argument about shrinking [itex]R_1[/itex]. My erroneous solution does not necessarily tend to zero when [itex]R_1[/itex], and therefore also [itex]r[/itex], do. If you take [itex]R_1 = r + a r^2[/itex] for example, where [itex]a[/itex] is a positive real number, then [itex]\frac{1}{r} - \frac{1}{R_1}[/itex] tends to [itex]a[/itex], if I calculated it right. It seems to me that this limit situation is ill defined:
jch1

jch, when r tends to zero, [itex]r^2[/itex] is far less than [itex]r[/itex], so again we have
[itex]R\approx r[/itex].

rude man, I don't know the right answer. I found the problem on some forum, but there was no solution. Anyway its good problem.
 
  • #24
SammyS
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The image charge does replace the conductor.

At locations external to the conducting sphere, the electric potential due to the image charge and the point charge, is the same as the electric potential due to the conductor and the point charge.

That is not the case interior to the outer surface of the sphere. At locations interior to the sphere, the electric potential is constant & the electric field is zero.
 
  • #25
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sammy, we don't have spherical shell, its thick shell. one image charge would not be sufficient...
 

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