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Homework Help: Electric Potential Due to Spherical Shells

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A spherical shell of radius R0 has a non-uniform surface charge density: η=η0*cos(2θ), where θ is the angle measured from the positive z axis and η0 is a constant. This shell is inserted into another spherical shell (this one has a volume) with its inner lip at radius R1 and outer lip at radius R2 and in the volume between there is a perfect conducting material (denoted σ→∞). this outer shell is electrically neutral.

    The two spheres are concentric.

    Except for the volume between r=R1 and r=R2, the whole space is characterized by the vacuum constants, ε0 and μ0

    What is the electric scalar potential everywhere?

    I attached a sketch of the problem.

    2. Relevant equations

    Laplace Equation in spherical coordinates with azimutal symmetry (∂/∂[itex]\varphi[/itex]=0):


    3. The attempt at a solution

    In the regions r<R0 and R0<r<R1 I write the general solution for the laplace equation:
    ∑(Anrn+Bnr-n-1)Pn(cosθ), n goes from 0 to ∞ and Pn() are lagandere polynomials.

    In the region r<R0 i set Bn=0 so that the solution wont "explode" when r→0. I "stitch" the two potentials (from either side) at r=R0 through the continuity of the tangential electric field and the discontinuity of the normal electric field due to η.

    Here i got stuck. I think that I need to find the boundary conditions at r=R1 and r=R2 but I cant find it.

    Any help would be appreciated,

    Attached Files:

  2. jcsd
  3. Dec 26, 2013 #2


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    The potential is continuous at the inner sphere, and as the outer sphere is conducting the potential there ...

    Everything outside of R1 is easy to solve.
  4. Dec 26, 2013 #3
    I understand that inside the volume of the outer sphere the potential is constant and equals to the potential at r=R2. but what is this potential?

    In the solution that was given to us, the teacher just applied gauss's law in integral form to the region r>R2, found the electric field and then the potential. then he found the potential at r=R2 and since the volume is perfectly conducting it is also the potential at r=R1 and from there you have a boundary condition to solve for the unknown coefficients of the laplace equation.
    I get all the proccess and it is pretty easy BUT in order to even start with applying gauss's law the teacher made the assumption that the electric field in the r>R2 region is fully spherically symmetrical meaning it has only a radial component.

    how can this be if the charge density is non uniform??
  5. Dec 26, 2013 #4


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    The potential at r=R1 is the same everywhere (and the electric field strength between R1 and R2 is zero), therefore the problem "outside" has the full spherical symmetry. Every non-uniformity of the charges at the inner sphere is cancelled by the non-uniform charge distribution at the inner surface of the conductor.

    That is your choice - the absolute value of the potential is arbitrary, as only potential differences have a physical meaning. It is convenient to have 0 as limit for r->infinity, but it is not necessary.
  6. Dec 26, 2013 #5
    well ill take it as it is but it still puzzles me how in a conductor one edge has non uniform distribution and at the outer edge the distribution is "suddenly" uniform. what prevents it from being also non uniform?
  7. Dec 26, 2013 #6


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    The conductivity of the material
  8. Dec 26, 2013 #7
    so why in the inner edge of the outer sphere the distribution is non uniform? its the same conductor having the same conductivity. shouldnt it be uniform just with the total charge on that edge the same as the total charge in the inner sphere?
  9. Dec 26, 2013 #8


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    As seen from the outside: otherwise the non-spherical contributions would not cancel.
    As seen from the inside: The non-uniform field between R0 and R1 leads to a non-uniform induced charge distribution at R1 (remember: charge density there is proportional to the electric field strength close to the surface).
  10. Dec 26, 2013 #9
    okay ill ponder on this a bit more on my own..

    thank you very much for your help and the extremely fast responses!
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