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What is the potential between 2 concentric spheres?

  1. Aug 19, 2014 #1

    Red

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    1. The problem statement, all variables and given/known data
    Consider two concentric spherical conducting shell. The inner sphere has radius r1, potential V1, while the outer sphere has radius r2, potential V2. Find the potential at the center of these two sphere, at r0=(r1+r2)/2.

    2. The attempt at a solution
    I tried to use method of images, by modelling there to be a charge q1 at the center of the inner sphere and q2 at the center of the outer sphere. Since both sphere are concentric, q1 and q2 are at the same location. So I attempt to add up q1 and q2 (which I now know is wrong), and then find the potential at r0. My answer is double the correct answer.
     
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  3. Aug 19, 2014 #2

    Zondrina

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    You can treat each concentric sphere like a point charge and calculate the potential of each one using ##V = k \frac{q}{r}##.

    You have to add the potentials at the point to get the net potential. Could you show your working for ##V_1## and ##V_2##?
     
  4. Aug 19, 2014 #3

    Red

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    Sure. I let V1=-kq1/r1, V2=-kq2/r2. I made both q the subject and sum them up, q0=q1+q2=(-r1V1-r2V2)/k. Then I proceed to find the potential at center of both sphere, V0=-kq0/r0=(r1V1+r2V2)/r0.
     
  5. Aug 19, 2014 #4

    Zondrina

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    So both charges are negative?

    How far away is the inner point charge from the point?

    How far away is the outer point charge? Those are the radial distances you should be concerned with.

    Then sum ##V_{net} = V_1 + V_2##.
     
  6. Aug 19, 2014 #5

    haruspex

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    Not so fast - you cannot treat a uniform shell of charge as a point charge for calculating potentials inside the shell.

    The potential at the outer shell is not solely due to the charge on the outer shell. There is a contribution from the inner shell.
     
  7. Aug 19, 2014 #6

    Red

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    Yes exactly, the potential at the outer shell is related to the charge that contribute to the inner shell. Haruspex how can I find the mean potential at the center of the two shells?
     
  8. Aug 19, 2014 #7

    haruspex

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    There may be a quicker way, but this way looks safe:

    Suppose the charges are q1, q2. Compute the potentials that should result. What equations does that give you?
     
  9. Aug 19, 2014 #8

    Red

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    I don't quite get you. If I sum up q1 and q2 then I get V0=(r1V1+r2V2)/r0.
     
  10. Aug 20, 2014 #9

    ehild

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    Assume that the charge on the inner shell is q1 and on the outer shell is q2. Applying Gauss' Law, you can determine the electric field E(r) between the shells (E1) and outside the bigger shell (E2).
    Remember that the electric field is gradient of the potential: you get the potential difference between two points if you integrate -E(r) between those points. Choose V(∞)=0 and integrate E2 between r2 and infinity. Integrate also E1 between r1 and r2. You get two equations which relate the unknown charges to the potentials V1 and V2. If you know the charges it is easy to get the potential at r=(r1+r2)/2

    ehild
     
  11. Aug 20, 2014 #10

    Red

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    Hi ehild, thank you for your response, however the charge on the shells are not known. What is known is that the outer shell has potential V2 and the inner shell has potential V1. How do I find the potential at the center between these two shells?
     
  12. Aug 20, 2014 #11

    haruspex

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    No, I mean compute the potentials that ought to result at the two shells. Equating those to the given potentials allows you to deduce the charges.
    Note that the two potentials are not simply q1/r1 and q2/r2, which you seem to be assuming.
     
  13. Aug 20, 2014 #12

    ehild

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    Find the relation between the charges and the potentials. From those, you can determine the charges.

    Supposing the inner shell has q1 charge. What is the electric field around it in terms of q1 if r1<r<r2? Apply Gauss' Law.

    ehild
     
    Last edited: Aug 20, 2014
  14. Aug 20, 2014 #13

    Red

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    Thank for your responses. I think there is an misunderstanding. Maybe I should ask the question in a slightly different way: A potential VH is applied between 2 spherical conducting shell. The inner sphere has radius r1, while the outer sphere has radius r2. What is the potential at the center of these two sphere, at r0=(r1+r2)/2?
     
  15. Aug 20, 2014 #14

    Orodruin

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    Assume that the shells have charges ##q_1## and ##q_2##, respectively. What would be the potential between the shells in this situation? (Noting that you are inside one of the shells.)
    If you have an expression for the potential given the two charges, then you can apply the boundary conditions to this, i.e., the potential at the outer shell has to be ##V_2## and the inner ##V_1##. This gives you two equations and two unknowns, so you will be able to deduce the charges and thus the potential. After this you should be able to simply plug in whatever ##r## you want into this.
     
  16. Aug 20, 2014 #15

    haruspex

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    It doesn't matter how the shells have the given potentials. It might as well be via an applied charge, so you can solve it the way ehild, Orodruin and I are suggesting. We're all telling you the same method, so how about trying it?
     
  17. Aug 20, 2014 #16

    ehild

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    You misunderstand the problem. The two shells are concentric, the radii are given, r1 and r2. The potential is given for both spheres; V1 and V2 . The problem asks the potential at distance ro=(r1+r2)/2 from the common centre of the spheres. In the centre, the potential is the same as on the inner sphere, that is V1.

    If you do not want to use charges, remember that the electric potential U obeys the Laplace equation at points where the charge density is zero. Because of the spherical symmetry of the problem, U depends only on the radius r, and [tex]\frac{1}{r^2}\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0[/tex] with the boundary conditions U(r1)=V1, U(r2)=V2.
    See http://en.wikipedia.org/wiki/Laplace_operator
    What is the solution for U(r) between the shells r1<r<r2?

    ehild
     
  18. Aug 20, 2014 #17

    Orodruin

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    Or, put in another way, as long as you are not putting charges between the shells, any charge configuration that fulfills the boundary conditions is going to give you the same potential. You might as well select a surface charge on the shells. Or you could assume a point charge in the common center of the spheres and a surface charge on a shell with radius ##R \gg r_2##. Both approaches will give you the same form for the potential between the shells, which can then be adjusted to accommodate the boundary conditions.

    The alternative is to do what ehild suggests and bite into the Laplace equation with spherical symmetry in spherical coordinates. This requires at least some knowledge about differential equations and how to solve them (in particular differential equations of Euler type).

    Note to self: You have been spending too much time on PF when you start writing ## instead of $ in your LaTeX source ...
     
  19. Aug 20, 2014 #18
    Potential in inner part of spear is always zero due to equal distribution of charge. ..Right?.....but outside of spear there is net potential ...okey so.

    Answer is....Potential in between spear is only due to inner spar....outer spear effect is zero.....But outside of whole spear there is net potential occur.......... So

    Answer:
    (1).inside the inner spear,potential is zero.
    (2).between two spear, potential is only due to inner spear.
    (3).outside the whole spear there is potential present due to both spears.
     
  20. Aug 20, 2014 #19

    ehild

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    The potential inside a conductor is constant.
    And what do you mean on spear? The problem is about concentric spherical shells.

    ehild
     
  21. Aug 20, 2014 #20

    Orodruin

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    If I am going to attempt a translation of DRU's post, I do believe "spear = sphere" and "potential = field" ...
     
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