Finding potential at the center of metal sphere

[issacnewton]

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 let's 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

 

[Liquidxlax]

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

 

[issacnewton]

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

 

[rude man]

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

 

[Liquidxlax]

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

 

[issacnewton]

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.

 

[rude man]

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.
.

 

[issacnewton]

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.

 

[rude man]

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.

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).

 

[issacnewton]

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 realize that I made a mistake
while typing the problem..

 

[rude man]

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.

 

[rude man]

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

 

[issacnewton]

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 ?

 

[rude man]

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.

 

[issacnewton]

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.

 

[rude man]

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!

 

[jch1]

(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

 

[issacnewton]

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 .

 

[rude man]

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.

 

[issacnewton]

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...

 

[jch1]

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 \frac{q}{4 \pi \epsilon_0 R_2}, as you claimed. If I rerun the last part of my argument with this new value, I get: U(0) = average of U on the inner surface = average of \frac{q}{4 \pi \epsilon_0 R_2} - U_q on the inner surface = average of \frac{q}{4 \pi \epsilon_0 R_2} on the inner surface - average of U_q on the inner surface = \frac{q}{4 \pi \epsilon_0 R_2} - \frac{q}{4 \pi \epsilon_0 R_1}. (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 = 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}, 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 R_1. My erroneous solution does not necessarily tend to zero when R_1, and therefore also r, do. If you take R_1 = r + a r^2 for example, where a is a positive real number, then \frac{1}{r} - \frac{1}{R_1} tends to a, 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 R_1 to O), 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

 

[rude man]

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 R_1 to O), 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!

 

[issacnewton]

This being said, I do not agree with your argument about shrinking R_1. My erroneous solution does not necessarily tend to zero when R_1, and therefore also r, do. If you take R_1 = r + a r^2 for example, where a is a positive real number, then \frac{1}{r} - \frac{1}{R_1} tends to a, if I calculated it right. It seems to me that this limit situation is ill defined:
jch1

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

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.

 

[SammyS]

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.

 

[issacnewton]

sammy, we don't have spherical shell, its thick shell. one image charge would not be sufficient...

 

[SammyS]

sammy, we don't have spherical shell, its thick shell. one image charge would not be sufficient...

Is the outer surface a sphere ?

 

[rude man]

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.

It's not a sphere, it's a spherical shell with thickness R₂ - R₁.

We're all having a tough time figuring out what the field is inside the shell. The problem is easily solved via image if q is outside the shell (r > R₂) but unfortunately someone stuck it inside ...

 

[SammyS]

It's not a sphere, it's a spherical shell with thickness R₂ - R₁.

We're all having a tough time figuring out what the field is inside the shell. The problem is easily solved via image if q is outside the shell (r > R₂) but unfortunately someone stuck it inside ...

Thanks for clearing that up for me.

The method of image charge, can give the potential inside the inner surface of the thick shell. Basically, the image charge method gives equi-potential surfaces which are spherical. Choose an image charge which makes the potential constant at r = R₁, the inner radius.

For r ≤ R₁, the potential is described by the equivalent system of the charge and image charge.

For R₁ ≤ r ≤ R₂, the potential is constant.

For r ≥ R₂, the potential is equivalent to the potential due to a charge of q located at r = 0 .

Of course, you have to decide on the location at which the potential is zero. A constant term may be added to the potential in any of the above regions so that the potential function is a continuous function of r.

 

[rude man]

Thanks for clearing that up for me.

The method of image charge, can give the potential inside the inner surface of the thick shell. Basically, the image charge method gives equi-potential surfaces which are spherical. Choose an image charge which makes the potential constant at r = R₁, the inner radius.

For r ≤ R₁, the potential is described by the equivalent system of the charge and image charge.

For R₁ ≤ r ≤ R₂, the potential is constant.

For r ≥ R₂, the potential is equivalent to the potential due to a charge of q located at r = 0 .

Thanks for your post. I agree with your statement for r < R₁ and we need to come up with that charge magnitude and location. This is of course how any image is created, by satisfying the boundary condition of equipotential somewhere.

I don't agree with your statement if r > R₂. I have previously addressed that situation in my post (#4) and am quite sure it's correct, since I lifted it straight out of the textbook I referenced. But, as IssacN pointed out later, q, and therefore r, are definitely inside the cavity, or r < R₁.

Bottom line, we need to find the location and magnitude of the image charge located somewhere! We need to recognize that that location could be inside the cavity, within the metal, or outside!

 

[SammyS]

...

I don't agree with your statement if r > R₂.
...

That's unfortunate.

Use Gauss's Law. It's really fairly simple to show that for an electrostatic condition, the charge distribution outside of a closed conductor is independent of any charge distribution inside, and furthermore depends only on the net charge inside, and of course on any external charges. In this case there are no external charges.

Now, if you want to include the charge, q, and the image charge in describing the potential external to the conductor, that needlessly complicates matters.

 

[rude man]

That's unfortunate.

Use Gauss's Law. It's really fairly simple to show that for an electrostatic condition, the charge distribution outside of a closed conductor is independent of any charge distribution inside, and furthermore depends only on the net charge inside, and of course on any external charges. In this case there are no external charges.

Now, if you want to include the charge, q, and the image charge in describing the potential external to the conductor, that needlessly complicates matters.

By definition, if r > R₂ the charge q does reside outside the shell, so there are "external charges"! Are you confusing r as the location of the potential instead of the location of q?

Once again: q is inside the cavity, so r < R₁. We don't need to debate the situation where r > R₂. However, as a side issue, if r > R₂ you should IMHO reconsider your statement.

 

[SammyS]

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 &lt; 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 let's 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

You have a conflict with your choice of variables.

Is the variable, r, the distance of an arbitrary point from the center of the spherical shell, or is r the distance that charge q is from the center of the spherical shell ?

Let's let the distance that q is from the center of the shell be a . You need to place the charge at a particular location, perhaps on the z-axis.

Your solution is correct for r ≥ R₂, and for R₁ ≤ r ≤ R₂, \displaystyle V(r)=\frac{1}{4\pi\epsilon_0}\frac{q}{R_2}\,, a constant value.

For r ≤ R₁, use the method of image charges. An image charge of Q, will be located along the z-axis, a distance, d, from the common center, with  \displaystyle \frac{Q}{q}=-\frac{R_1}{a}\,,  and  \displaystyle a\cdot d={R_1}^2\,.

The charge, q, along with the image charge, Q, will give you a potential of zero at r = R₁, so add to that the constant potential you previously obtained at r = R₁, which is the same as the potential at r = R₂ .

 

[rude man]

There is no conflict of variable choice. r was always defined as the location of q, with
r < R₁.

The one and only point we're interested in as far as potential is concerned is the center of the spherical shell. There is no need to introduce further variables. There is q, R₁, R₂ and r. That is all ye know on Earth, and all ye need to know.

Less facetiously: thanks for coming up with the image for r < R1. I hope to check it but it looks right. Which solves the problem once and for all if so.

For r > R2 the image's magnitude and location are per my post (#4).

 

[SammyS]

There is no conflict of variable choice. r was always defined as the location of q, with
r < R₁.

The one and only point we're interested in as far as potential is concerned is the center of the spherical shell. There is no need to introduce further variables. There is q, R₁, R₂ and r. That is all ye know on Earth, and all ye need to know.

Less facetiously: thanks for coming up with the image for r < R1. I hope to check it but it looks right. Which solves the problem once and for all if so.

For r > R2 the image's magnitude and location are per my post (#4).

If r is the location of the point charge and you're only interested in the potential at r, then the answer is easy.

The potential at the location of the point charge is ±∞ , depending upon the sign of the charge.

I hope that is not what OP was interested in.

 

[rude man]

Look again at what I said. To repeat verbatim et litteratim: "The one and only point we're interested in as far as potential is concerned is the center of the spherical shell."

That point is not r. That point is O(0,0,0) if we center the shell at the origin. r is the distance from the origin to q. q cannot sit at O.

 

[rude man]

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 ?

Going back to your question here, I think I can give you a better answer.

Put a Gaussian spherical surface around O, radius between R1 and R2. The E field everywhere over this surface is zero since it's in a metallic environment. So the charges at the outside surface (at R2) do not experience a radial force. That means that the only force exetrted on them is that due to their neighbors, i.e. all on the surface R2, and that obviously will distribute the charges so that there is uniform distance between them, in other words, uniform charge density.

 

[SammyS]

...

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

I'm sorry that I misinterpreted and misread the problem,

Thanks to rude man for pointing out the details of the problem.

I fully agree with the result you have in the above quote.

The potential at the center of the spherical shell is \displaystyle V_{\text{at the center}}=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}-\frac{q}{R_1} \right)\, relative to the inner surface of the shell.

The inner surface of the shell is at the same potential as the entire shell, including the outer surface, which is at a potential of \displaystyle V(r)=\frac{1}{4\pi\epsilon_0}\frac{q}{R_2}\, assuming the potential at infinity to be zero.

 

[rude man]

Good work, everybody, and congrats to IssacN who got it right the first time! We are all on the same page, finally!

 

[ehild]

I have problems with the method of image charges. Replacing the metal surface by an image charge works perfectly when the real charge is outside a grounded sphere. When a=R1 the potential of the charge-image system will return zero, which has sense as an opposite charge from the Earth would cancel its effect.
In case of a stand-alone metal sphere a charge Q' equal and opposite to Q has to be placed into the centre, and the potential outside the sphere is obtained with that three-point-charge system.

Here the problem is that the charge is placed inside the sphere. Both the magnitude and distance of the image charge is obtained with respect to the inner sphere when it is at zero potential. But then a charge of -Q has to be added somewhere outside, but where? Without doing so, we added a charge Q to the system, so it is not equivalent to the original one.

Does the potential inside the sphere really tend to q/R2 when r approximates R1? Which would mean putting that extra charge Q' at infinity...



ehild

 

[issacnewton]

Wow,

So many posts. I only got one email notification for all these posts. Anyway great discussion.
I have some problems with the method of images presented. What about the equipotential
spherical surfaces between R₁ and R₂. Don't we need images charges
for them. Maybe that's what ehild is trying to say. My understanding of method of images
is from Feynman's lectures. I couldn't understand Griffiths properly here.

Sammy, sorry for the confusion you had to go through. I didn't type the problem correctly. r is
a fixed variable here as I pointed out later.

rude man, thanks for the explanation in post # 36. After reading some Feynman, that's the
understanding I developed too. I am sure there is rigorous mathematical reasoning for that. But
for basic physics the argument given by you should suffice I guess.

 

[rude man]

ehild, there is no doubt left that IsaccN's original anwer is correct. Now, I haven't verified that that answer is obtainable by the magnitude and location of SammyS's image charge, but if it is then what he did was ipso facto correct.

I would sum up the situation, which as I say produces the answer that IssacN, SammyS and I agree on, as follows:

Potential at R₂ is kq/R₂ by Gauss, as SammyS pointed out.

Then, V(R₁) = V(R₂) 'cause we're going thru metal.

Inside the cavity, while we have no clue as to the exact distribution of the negative induced charges at the cavity wall (R₁), we do know that all the charges have to add up to q, again by Gauss, and since they're all equidistant from O, the potential due to all those charges at O is just -kq/R₁. It's of course negative since the induced charges are also.

Finally, obviously the potential at O due to q itself, located at r, is kq/r.
So add them all up and you get IssacN's original answer!

 

[SammyS]

...

Does the potential inside the sphere really tend to q/R2 when r approximates R1? Which would mean putting that extra charge Q' at infinity...



ehild

Hello ehild !

As r → R₁, the potential inside the sphere does tend to q/R₂, but the position of the image charge tends to the position of the charge, i.e. its distance from the center of the sphere approaches R₁. As for the amount of charge of the image charge, Q: Q → -q .

 

[ehild]

Hello ehild !

As r → R₁, the potential inside the sphere does tend to q/R₂, but the position of the image charge tends to the position of the charge, i.e. its distance from the center of the sphere approaches R₁. As for the amount of charge of the image charge, Q: Q → -q .

That I knew Sammy. My problem is not the image charge but the position of the charge which compensates the image charge in the method of images.
Now I think that the space between the inner sphere and infinity can be filled with a metal, and then the sphere can be taken "grounded", no need of compensating charge. But the "ground" is at potential q/R2.

ehild

 

[ehild]

ehild, there is no doubt left that IsaccN's original anwer is correct. Now, I haven't verified that that answer is obtainable by the magnitude and location of SammyS's image charge, but if it is then what he did was ipso facto correct.

rude man,

I also think IsaccN's solution correct. I have problems with the method of images.

ehild

 

[ehild]

Wow,

So many posts. I only got one email notification for all these posts. Anyway great discussion.
I have some problems with the method of images presented. What about the equipotential
spherical surfaces between R₁ and R₂.

Yours was a great solution, congratulation!
There is a theorem in Electrostatics that the inside of an equipotential surface can be filled with metal, it does not change anything outside. The beauty of this problem was that the inside was outside. I think it is only the inner surface of the shell for the image charge you need to take into account. The inside of the shell is not influenced by the outer surface except the additive constant of the potential.

ehild

 

[rude man]

rude man,

i also think isaccn's solution correct. I have problems with the method of images.

Ehild

Well, did you "try it on", meaning did you use SammyS's image charge magnitude and location and find that it did not give what we all agree is the correct answer? I was going to do that myself but I'm too lazy ...

SammyS, did you? I would have expected you did, seeing as you agree with IssacN's answer.

 

[ehild]

Well, did you "try it on", meaning did you use SammyS's image charge magnitude and location and find that it did not give what we all agree is the correct answer? I was going to do that myself but I'm too lazy ...

I have done it, but the method works for a grounded metal surface, at zero potential, and results with the magnitude and distance for the image charge cited by SammyS. Assuming grounded shell, the potential at the middle is kq/r - kq/R1. Saying that it is with respect to the potential of the inner surface which is at kq/R2 potential with respect to infinity, we add that term to U(0). The result is the same as IsaacN's. I have doubts concerning that additive term.
The potential was assumed to be k*q/distance which means zero potential at infinity. If the zero of the potential is at R1, the potential is not k*q/distance.

In case the metal is not grounded and the original charge is outside the sphere my book says to add a charge equal in magnitude with Q and of opposite sign to the centre, to preserve charge.
My problem is that extra charge, when the original charge is inside the shell.

It is impossible to find an "image charge" if the sphere is not at zero potential with respect to infinity.

ehild

 

[issacnewton]

There is a theorem in Electrostatics that the inside of an equipotential surface can be filled with metal, it does not change anything outside. The beauty of this problem was that the inside was outside. I think it is only the inner surface of the shell for the image charge you need to take into account. The inside of the shell is not influenced by the outer surface except the additive constant of the potential.

ehild

What is this theorem ? Any name ?

It seems that image approach does not work in all cases in this particular problem. So my
method is probably better here. We don't have to worry about the images...

 

[rude man]

ehild, potential is classically defined as zero at infinity. It's always V/r in my book (Halliday & Resnick). So it seems that SammyS has determined the image correctly.

IssacN, I think imaging is extremely important and good to understand, even though I agree that your reasoning got us the answer pronto. I myself did not follow your reasoning 100% and had to do some hard thinking and stumbling along the way before agreeing with your answer, as you know. I wasn't totally convinced until I wrote my final post on the subject, "filling in the blanks" for myself. Both approaches ought to be given full credit.

 

[ehild]

What is this theorem ? Any name ?

I do not know the name if it has got any. But think: the metal surface is equipotential. The grad U lines (electric field lines) are perpendicular to the metal surface just like to the equipotential surfaces. So filling the inside of a closed equipotential surface with a metal does not change the field outside.

It seems that image approach does not work in all cases in this particular problem. So my
method is probably better here. We don't have to worry about the images...

You applied the method replacing the metal with the surface charge distribution formed on it. And the method worked well, as the potential in the centre was asked. To determine the field at any point inside the cavity, you would need the distribution of the surface charges on the inner surface. With the method of image charges, the calculation is easy: you get the inner electric field as that of two point charges - the real one and the image.


ehild