Finding potential at the center of metal sphere

[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

 

[ehild]

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

 

[rude man]

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.

 

[issacnewton]

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.

 

[rude man]

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.

 

[ehild]

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

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.

ehild

 

[ehild]

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

 

[rude man]

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.

 

[issacnewton]

As I think about this problem, I think the image approach will work even in situations where
the inner spherical cavity of radius R_1 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.

 

[issacnewton]

rude man or ehild, can you confirm my opinion in post # 58

 

[ehild]

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