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Homework Help: Point Charge in a Glass Sphere

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    There is a point charge Q = 4π nC inside a glass sphere which has the radius 1 m and the dielectric constant εr=10. This sphere is enclosed in a hollow metal sphere with an inner radius of 2 m and a thickness of 0.5 m.

    The task is to find the amount of the electric field strength in the glass sphere, in the gap between glass and metal, in the metal layer and on the outside.

    2. Relevant equations
    Gauss law of electrostatics.

    3. The attempt at a solution
    As far as I know, the field outside of the metal sphere should be as if the sphere would be a point charge, i.e.E(r)=Q/(4πε0r2). But I am a bit lost at the other parts. Is the field in the glass sphere for instance simply: E(r)=Q/(4π εrε0r2)?
  2. jcsd
  3. Oct 13, 2011 #2


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    Correct. :approve: (Of course this is only correct due to the spherical symmetry of the whole thing. If it were any other shape, the answer would be different.)
    Yes, the equation for inside the glass sphere is also correct. :approve:

    You should be able to use the standard version of Gauss' law for the region (gap) between the glass sphere and hollow metal shell (εr = 1 in this region).

    The electric field within a conductor is special [Edit: by that I mean within the conducting material itself]. But I'll leave that for you to look up. :smile:
    Last edited: Oct 13, 2011
  4. Oct 14, 2011 #3
    OK, thanks :)! In the metal, there should be a shift of charges due to the field from the inside, so that the inside of the metal layer is free of the electric field, i.e. |E(r)|=0?

    One last pointer on the fields inside though, please. Do I need to incoporate the radius of the glass sphere into the calculation somehow? Or can I just say that below r0 (which is the radius of the glass sphere, the inner radius of the metal was denoted as r1 and the thickness was given as d, so the outer surface of the metal sphere is at r1+d), the field is E(r<r0)=Q/(4π*ε0r*r2), then in the air gap, the field is E(r0<r<r1)=Q/(4π*ε0*r2)?

    And accordingly, E(r>r1+d)=Q/(4π*ε0*r2)?
  5. Oct 14, 2011 #4


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    Yes that's right. The (static) electric field within a conductor is always zero. Things get more complicated when you move on to electrodynamics. But with static charges, the field within a conducting material is always zero. :approve:
    The latter looks good to me. :approve:
    Yes, that's right! :approve: And that's for the same reason that you can say E = Q/(4π*ε0*r2) outside of the metal shell too. As long as there is perfect spherical symmetry, it doesn't matter what is on the outside of the Gaussian surface. All that matters is the total charge within.
    Yes. :approve: Good work.
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