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Two concentric conducting esferical shells connected by thin wire

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider that the electric field is proportional to q r-2-d[itex]\hat{r}[/itex], where q is magnitude of a point charge and d<<1.

    Two concentric conducting esferical shells for radii a and b (a>b) connected by thin wire. The external shell has a total charge qa. Prove that the charge of the smaller shell is:

    qb=- [itex]\frac{q_{a}d}{2(a-b)}[/itex]{2bln(2a)-(a+b)ln(a+b)+(a-b)ln(a-b)} + O(d2)

    2. Relevant equations
    [itex]\vec{E}[/itex] [itex]\propto[/itex] r-2-d[itex]\hat{r}[/itex]


    3. The attempt at a solution

    I just can't imagine how a charge would be localized in the smaller shell as I always took for granted that charges would place themselves on the external shell.
     
  2. jcsd
  3. Apr 3, 2013 #2

    TSny

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    Hello and welcome to PF!

    If the electric field of a point charge does not decrease as the inverse square of the distance, then a uniformly charged spherical surface will have a non-zero electric field in the volume enclosed by the surface. For example, if the parameter d > 0, then there will be a radially inward pointing electric field inside the sphere. If d < 0, there will be a radially outward pointing field. Only at the center will the field be zero.

    Think about what that implies for the electric potential inside the sphere.

    Then imagine two concentric spherical conducting surfaces such that the outer conductor is charged but the inner conductor is not charged. Would there be a potential difference between the two conductors? If so, would connecting a wire between the spheres cause the inner sphere to acquire a net charge?
     
  4. Apr 3, 2013 #3
    Thanks for your answer! It is really the lack of manipulation outside the scope of Maxwell ED that makes it difficult. I took for granted that, as they were connected by a wire, the potential would be the same.

    I'll try to solve again and I'll let you know if I did it!

    Thanks, again!
     
  5. Apr 3, 2013 #4

    TSny

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    That's right, without the inverse square field of a point charge then you don't have some of the nice things like Gauss' law, you can't replace a uniform sphere of charge by a point charge, etc.
    Yes, that's the correct idea. Connecting the spheres with a wire will cause charge to flow between the spheres until the potential is the same for each sphere.
     
  6. Apr 14, 2014 #5
    Hi, I am trying to find this answer, any news on this? I struggled, tried to use Gauss' law with the new E field, but it seems that the charges in both sides of equation cancel...
     
  7. Apr 14, 2014 #6
    That's a nice way to experimentally test the inverse square law which indirectly measures the photon mass.
     
  8. Apr 14, 2014 #7
    Hi dauto, yes actually there is a publication on this (doi:10.1103/PhysRevLett.26.721). But this is a Panofsky & Phillips problem (1.8). I tried a lot of things, and indeed this is a really hard situation, since we are all used to deal with inverse square law behavior. Let me know if you find something. thanks
     
  9. Apr 15, 2014 #8

    TSny

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    Gauss' law is not valid except for the specific case of inverse square behavior of the field of a point charge (d = 0).

    I was able to get the result by brute force by finding the potential, V, of a uniformly charged shell at arbitrary points inside and outside the shell in terms of q, r, and d. Here, r is the distance from the center of the sphere. The result can then be approximated to first order in d.

    In this problem, we have two charged spherical shells. Requiring the total potential to be the same at the surface of each shell gave the result.

    It was somewhat tedious, but not as bad as I thought it might be. I would appreciate knowing of a better method.
     
  10. Apr 15, 2014 #9
    I don't think there is another method. That's the way to do that problem.
     
  11. Apr 30, 2014 #10
    Hi guys,

    Well I don't understand what you mean by brute force. Any clue? I would like to try to solve it by myself. Also, it sounds a little bit strange that, well, the Gauss' law is not valid in this case, but the assumption of the *same* potential when the spheres are connected still is...any comments?

    Well I have tought a little bit and I assume that by "brute force" you mean:
    [itex]V_{A}-V_{B}=-\int_{B}^{A}\vec{E}\cdot d\vec{l}[/itex]. Is that right?
     
    Last edited: Apr 30, 2014
  12. Apr 30, 2014 #11

    TSny

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    From the given expression for E of a point charge, find an expression for the potential V(r) of a point charge.

    Then consider a uniformly charged spherical shell. By integration, find the potential at points inside and outside the shell. Then approximate the result for d << 1 keeping terms only through first order in d.

    Next consider two concentric spherical shells with the outer shell having charge qa and the inner shell having charge qb. By requiring the total potential at the surface of the outer shell (due to both shells) to equal the total potential at the surface of the inner shell, you can derive the relation between qb and qa.
     
  13. Apr 30, 2014 #12

    TSny

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    Yes, use this to find V(r) of a single point charge. You can let either point A or point B be a point at infinity and take V = 0 at infinity.
     
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