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Three concentric spherical conductors -- Find the potential

  1. May 4, 2016 #1
    1. The problem statement, all variables and given/known data

    I am not sure whether to put this in the introductory level or advanced. It seems to be relatively introductory in an electromagnetism course.

    A spherical conductor of radius ##a## carries a charge ##q##. It is situated inside a concentric spherical conducting shell of inner radius ##b## and outer radius ##c##, ##a<b<c##, which carries a charge ##Q##. These two are in turn surrounded by a spherical conductor radius ##R##, which is earthed. Find the potentials at all points in the regions between these conductors.


    3. The attempt at a solution

    Concentric conducting spheres 2.PNG

    I thought there might be three different ways I could do this, but it seems for one I get significantly different answers and I would like to know why (also I am not sure if any of my answers are correct). I will use the region ##c<r<R## to show each method.

    Method One: Compute the electric field in the region using Gauss' law and then find the potential difference between two points using the path integral ##\phi_{ab}=\int_a^b\vec{E}\cdot\vec{dr}##.

    In the region ##c<r<R## the electric field is: ##\vec{E}=\frac{Q+q}{4\pi\varepsilon_0r^2}\hat{r}##. Now the potential difference between a point r (c<r<R) and R is: $$\phi_{rR}=\int_r^R\frac{(Q+q)}{4\pi\varepsilon_0 r'^2}dr'=\frac{(Q+q)}{4\pi\varepsilon_0}\big(\frac{1}{r}-\frac{1}{R}\big)$$.

    By some luck this works in that at r=R the potential is 0. But this is only because it is giving the potential difference between the same point. For this reason, and because this gives the potential relative to R, I don't think this is the answer the question seeks. (Consider if the potential at R was set to, say, 5. There are no boundary conditions I have the opportunity to use). Do you agree?

    Method Two:
    Compute the electric field in the region using Gauss' law and then compute the potential using ##\vec{E}=-\vec{\nabla}\phi##.

    Assuming zenith and azimuthal symmetry then ##-\vec{\nabla}\phi## in spherical coordinates reduces to ##-\frac{\partial \phi}{\partial r}\hat{r}##. Which implies ##E(r)=-\frac{\partial \phi}{\partial r}##. So we have that in the region ##c<r\leq R## where the electric field is the same is mentioned before : ##\phi(r)=\frac{(Q+q)}{4\pi\varepsilon_0 r} +m## for some constant m. Using the boundary condition ##\phi(R)=0##, I get that ##m =\frac{-(Q+q)}{4\pi\varepsilon_0}## so that ##\phi(r)=\frac{(Q+q)}{4\pi\varepsilon_0}\big(\frac{1}{r} -\frac{1}{R}\big)##.

    This agrees with method one, but not by luck. For this reason I think that this method is probably the most correct. What do you think?

    Method Three: Solve laplaces equation for the potential in the region ##c<r<R## with the boundary conditions ##\phi(R)=0## and ##\phi(c)=V_c##.

    Assuming spherical symmetry and no angular dependence, ##\nabla^2\phi=0\implies \frac{1}{r^2}\frac{\partial}{\partial r}\big(r^2\frac{\partial \phi}{\partial r}\big)=0\implies \phi=\frac{-m}{r}+n## for some constants m and n. Using the boundary conditions I conclude that ##\phi=\frac{cV_c}{c-R}\big(1-\frac{R}{r}\big)##.

    This potential works in that when r=R the potential is zero, and at r=c the potential equals Vc=Vb, but only because I constructed it so. The problem I have with this method is that 1. The answer appears to be significantly different to that of method 2 (which I believe to be correct) and 2. There seems to be some ambiguity in what Vc, Vb and Va is, and it doesn't seem that I can calculate them using this same method, but maybe if I used method two. Or maybe this method is completely wrong. What do you think? Why does method 2 and 3 seemingly give different answers?
     
  2. jcsd
  3. May 4, 2016 #2

    haruspex

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    Your method 1 is fine, and is what I would use. There is no problem taking the potential at R to be zero since you are told it is earthed.
    I see no evidence that method 3 produces a different answer. You would need to plug in the value of Vc to demonstrate that.
     
  4. May 5, 2016 #3
    After some extra thought I agree that method three does not give a different answer.

    I have followed method one all the way through to answer the question and my final answer would be this:

    ##\phi(r)=
    \begin{cases}
    V_A, & 0\leq r\leq a\\
    \frac{q}{4\pi\varepsilon_0}\big(\frac{1}{r}-\frac{1}{b}\big), & a<r<b\\
    V_b=V_c, & b\leq r \leq c\\
    \frac{Q+q}{4\pi\varepsilon_0}\big(\frac{1}{r}-\frac{1}{R}\big), & c<r \leq R\\
    \frac{Q+q}{4\pi\varepsilon_0}\big(\frac{1}{R}-\frac{1}{r}\big), &R\leq r
    \end{cases}##

    Which would look something like this:
    spherical conductor potential graph.png

    Does this seem acceptable/correct to you? Also just for future reference, is method two also correct? Thanks.
     
  5. May 5, 2016 #4

    rude man

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    You haven't addressed the region a < r < b at all. When you do, be sure to take into account the fact that the E field is zero for b < r < c. I would stick with method 1 since it's the easiest, but you're wise to pursue alternative methods. Keep at it until you get the same answers for all of them!

    EDIT: OK, I didn't see your last post. Looks OK.
     
  6. May 5, 2016 #5

    haruspex

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    I see a couple of problems in your answers.
    None of the answers should reference VA etc., they should all be in terms of the charges and the radii.
    The potential at infinity should be zero.
     
  7. May 5, 2016 #6

    rude man

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    I agree. I was very sloppy.
    I will reply to post #5 accordingly.
    I ignored the graph altogether.
     
  8. May 5, 2016 #7

    haruspex

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    I would let pondzo try to fix the errors first.
     
  9. May 5, 2016 #8

    rude man

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    Not possible if potential at R is to be zero.
    The solution to Laplace's equation here is
    Φ(r) = k1 + k2/r, k1 and k2 constants. Clearly k1 and k2 can't both be zero yet they would have to be if Φ(∞) = Φ(R) = 0.
     
  10. May 5, 2016 #9

    rude man

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    Yes, but this is a nontrivial problem - see my remark in post #8.
     
  11. May 5, 2016 #10

    haruspex

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    Why not?
     
  12. May 5, 2016 #11

    rude man

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    Did you look at my post #8? I tried to explain ..
     
  13. May 5, 2016 #12

    rude man

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    Let me say at this point that I would approach this problem by solving the Laplacian ODE , viz. ∇2(Φ) = 0.
    The solution is very simple in this case thanks to spherical symmetry, but you must solve it in 2 parts, i.e. the regions a to b and c to R, since the boundary conditions differ for them. Of course you then also know Φ in the metal regions.

    EDIT: And one big hint: one of the boundary conditions is on Φ, the other on dΦ/dr. But what is dΦ/dr?
     
    Last edited: May 5, 2016
  14. May 5, 2016 #13
    How would I Find the values of ##V_A## etc? I thought I might be able to via ##C=\frac{Q}{V}## but it seems like a circular problem. I thought that ##V_A,V_B=V_C## could be set to anything in a physical setup, and post #3 would be true, but judging by the replies it seems not.

    This is what I was thinking. Once I drew the graph of my solution for the potential, I thought it was odd that the potential went off to infinity as r went to infinity, but I only thought that because the normal convention is to set ##\phi(\infty)=0##. But in this case it is different is it not? Since we are Defining ##\phi(R)=0##. We can't have both equal to zero, or can we?

    This is partially why I wanted to use method two/three, because it seemed more 'surefire' and less room for confusion. I will refer back to my notes when I have the chance, but I think I may have got a more satisfactory answer using method two which wasn't in terms of ##V_A## etc.
     
  15. May 5, 2016 #14

    rude man

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    [

    I can now say that your method 2 is correct and the preferred way to go. Method 3 is obviously incorrect. I think method 1 may be OK but you haven't done the other region. I would stick to method 2 for both regions which was )and remains) my recommendation in post 12. Can't go wrong with old man Laplace!
     
  16. May 5, 2016 #15

    rude man

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    Right!
    You're doing fine, but let's also see the remaining part of the problem! :smile:
     
  17. May 5, 2016 #16

    haruspex

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    In post #8 you did not explain why k1 and k2 cannot both be zero. You merely stated it as though obvious.
    I don't see how there can be a field outside a closed earthed conductor (a Faraday cage).
     
  18. May 5, 2016 #17

    rude man

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    Upon further thought I think I agree with you. A test charge moved from infinity to the outer shell would require zero work, so the field for r > R is indeed zero. Thanks for your post and my apologies to the OP.

    As for the inside two fields I would proceed as I suggested.
     
  19. May 5, 2016 #18
    Haruspex could you please advise me on how you might go about the problem? Which approach would you take in order to find ##V_A## etc?

    So I had a look at my notes, and by using method two I concluded that:

    ##\text{ For } r\leq a, ~\phi(r)=\frac{Q+q}{4\pi\varepsilon_0}(\frac{1}{c}-\frac{1}{R})+\frac{q}{4\pi\varepsilon_0}(\frac{1}{a} -\frac{1}{b})##
    ##\text{ For } a<r<b ,~\phi(r)=\frac{Q+q}{4\pi\varepsilon_0}(\frac{1}{c}-\frac{1}{R})+\frac{q}{4\pi\varepsilon_0}(\frac{1}{r} -\frac{1}{b}) ##
    ##\text{ For } b\leq r\leq c, ~\phi(r)=\frac{Q+q}{4\pi\varepsilon_0}(\frac{1}{c}-\frac{1}{R})##
    ##\text{ For } c<r<R, ~\phi(r) =\frac{Q+q}{4\pi\varepsilon_0}(\frac{1}{r}-\frac{1}{R})##
    As for ##r\geq R## i'm not entirely how to sure to deal with it. It seems as though you are both saying that the electric field is zero outside the earthed conductor (a 'Faraday cage'), which means the potential must be constant, but how can I get it in terms of the charges held on the other conductors?

    To be completely honest, I'm now a little confused as to how answer/approach this problem. If I could have some guidance towards how you both might approach this problem.

    I'm sorry but I don't know why method three is obviously incorrect. And then you go on to say that you can never go wrong with Laplace, but that was the whole idea of method three. Maybe I am misinterpreting you.
     
  20. May 5, 2016 #19

    haruspex

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    I used method 1. Figure out the field in each region, use that to find the potential differences, then find the central potential that gives a potential 0 at R.

    Using that, I get the same answers as you for the potentials.
     
  21. May 5, 2016 #20

    rude man

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    Well, the potential is zero at r = R and there is no E field anywhere between infinity and R, so what does that tell you about the potential anywhere r > R?
    Method 3 is incorrect because you expressed your results in terms of Va Vb Vc etc rather than q and Q.

    What I got wrong is the boundary value on the r derivative of Φ(r), which of course is -E(r), at r = R. I assumed the E field at that point was k(q+Q)/R2 but that was wrong. E(r) = 0 for all r >= R. And the proof of that is what I said in my previous post, which is that there is no change in potential in going from r = ∞ to r = R, so the E field must be net zero from r = ∞ to r = R since the electrostatic field is conservative. If you use the correct boundary value of Φ(R) the Laplacian equation solves to the correct value in all 3 nonmetallic regions..

    At this point I suggest you stick with haruspex's way but it would be very instructive also to solve the problem with ∇2Φ = 0 as you have already attempted. Just use the correct boundary values for Φ for all three nonmetallic regions. The solution is always Φ(r) = k1/r + k2, k1 and k2 constants. These problems always have the boundary values as traps as I have fallen into!
     
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