# Finding potential at the center of metal sphere

1. ### issacnewton

621
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 $R_1$ and $R_2$ respectively.
Find the potential at the point O if $R_1 < R_2$.

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

$$V(r)=\frac{1}{4\pi\epsilon_o}\frac{q}{r}$$

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

$$V(R_2)=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}$$

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

2. ### Liquidxlax

322
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. ### issacnewton

621
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

$$V_{inner}=\frac{1}{4\pi\epsilon_o}\int \frac{dq}{r}$$

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

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

$$V_{inner}=\frac{1}{4\pi\epsilon_o}\frac{-q}{R_1}$$

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

$$V_{outer}=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}$$

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

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

I took some efforts to get this....
thanks

4. ### rude man

5,663
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

Last edited: Mar 6, 2012
5. ### Liquidxlax

322
that is why i suggested the "legendre" method because that allows for the R=0 case

6. ### issacnewton

621
If we consider the limit of $R_1$ 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

5,663
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. ### issacnewton

621
But, when you reduce $R_1$, you have to reduce r as well, otherwise you are
changing the conditions of the problem. So, when both $R_1$ and r go to zero,
$R_1 \approx r$ , so the first two terms cancel and you do get the potential
of the charge q on a metal sphere.

9. ### rude man

5,663
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).

Last edited: Mar 7, 2012
10. ### issacnewton

621
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 $r < R_1$. 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

5,663
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

5,663
No, they're not the same. You have the term kq/R1
wheras I have the term kq/(R1 + r).

13. ### issacnewton

621
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

5,663
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. ### issacnewton

621
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

5,663
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. ### jch1

2
(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 $$\frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R_1} \right).$$ This is rude man's first answer, except that I have $R_1$ where he has $R_2.$ My reasoning is as follows: you can write the potential inside the sphere as $U_q + U$, 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 $U_q$ is singular where the point charge is located, $U$ 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 $U = -U_q$ on the inner surface. So now you have an ordinary Dirichlet boundary value problem for $U$. 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 $U_q(0) + U(0)$, 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, $U = -U_q$ on the inner surface, so the average of $U$ is minus the average of $U_q$. To do the calculation it is best to use spherical coordinates, with the point charge on the zenith axis. Then $$U_q(R_1, \theta, \phi) = \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}}$$ (I used the cosine rule for this), and so the average value is $$\frac{1}{4 \pi R_1^2} \int_{\textrm{sphere}} U_q \, dS = \frac{q}{16 \pi^2 \epsilon_0} \int_{\theta = 0}^\pi \int_{\phi = 0}^{2 \pi} \frac{\sin \theta}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}} \, d\theta \, d\phi$$ $$= \frac{q}{8 \pi \epsilon_0} \int_0^\pi \frac{\sin \theta}{\sqrt{R_1^2 - 2 r R_1 \cos \theta + r^2}} \, d\theta = \frac{q}{16 \pi \epsilon_0 r R_1} \int_{(R-r)^2}^{(R+r)^2} \frac{du}{\sqrt{u}} = \frac{q}{4 \pi \epsilon_0 R_1}.$$ Hence, the potential at the center is $$\frac{q}{4 \pi \epsilon_0 r} - \frac{q}{4 \pi \epsilon_0 R_1} = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R_1} \right).$$

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

Last edited: Mar 8, 2012
18. ### issacnewton

621
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 $r$ and
$R_1$ shrink to zero, $R_1\approx r$ and then your solution tends to zero. But what should happen is that potential should tend to

$$\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}$$

because, there is charge q distributed on the outer surface. And my solution achieves that. Your doesn't .

19. ### rude man

5,663
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. ### issacnewton

621
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...

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