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Electric Field of Concentric Spheres and Opposite Charges

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    The figure to the right shows two concentric spheres made from insulators. One has radius and the other has radius R1, and the other has radius R2. The inner sphere has a positive charge density, +ρ, while the insulator region between the inner and outer spheres has a negative charge density, −ρ.

    IWL71Lh.png


    2. Relevant equations
    Gauss' Law

    3. The attempt at a solution
    a. I wrote expressions that represented each portion's charge densities, and set them equal to each other because the magnitudes of the charge densities are the same; I got 2.

    b. I used a Gaussian sphere with radius r and enclosed a portion of the smaller sphere; solved

    c. I used a Gaussian sphere with radius R2>r>R1. I expressed the charge of the small sphere as (4/3)π(R1)ρ. The other portion would be a sphere with radius r with a spherical cavity with radius R1 and charge density -ρ. I expressed its charge as (4/3)π(r^3-R1^3)(-ρ). I added these two together to get the enclosed charge, but I'm not sure how to express the flux in terms of E and dA. I know the negatively charged portion would have field lines entering the surface, and thus leading to a negative dot product, but would the positively charged sphere have an effect as well? Not sure what to do from here.

    Any help would be appreciated
     
  2. jcsd
  3. Sep 18, 2015 #2

    haruspex

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    You seem to be almost there. Having found the net charge inside the sphere radius r, R1<r<R2, isn't writing down the field at r immediate?
     
  4. Sep 18, 2015 #3
    Would it be the same as the electric field of a point charge?
     
  5. Sep 18, 2015 #4

    rude man

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    The formulation of this problem is terrible.

    There is one sphere, the one with R = R1, then there is a spherical shell from r = R1 to R = R2. There are not two shells, nor an "insulating layer" between them.

    Anyway, you're well on your way. I also don't see why it isn't immedialtely apparent to you what the E field has to be for R1 < r < R2. Just use Gauss's law! ε∫∫E⋅dA = Qfree inside the surface! It doesn't matter that some of the charge inside is + and some of it -, as long as all charge is distributed symmetrically with respect to the spherical coordinates φ and θ, which they are.
     
    Last edited: Sep 19, 2015
  6. Sep 19, 2015 #5

    haruspex

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    Yes, that's the neat thing about charges with a spherically symmetric distribution.
     
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