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Potential in a sphere in a dielectric

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    A solid conducting sphere of radius R and carrying charge +q is embedded in an electrically neutral nonconducting spherical shell of inner radius R and outer radius 2 R . The material of which the shell is made has a dielectric constant of 3.0.

    Relative to a potential of zero at infinity, what is the potential at the center of the conducting sphere?

    2. Relevant equations

    V = ∫Edr = kq/r + C

    Gauss's law in dielectrics: ∫KEdA = q/ε0

    3. The attempt at a solution

    I've made a few assumptions.

    The field in a conducting sphere is zero, which means the potential in a conducting sphere is constant. Looking at just the sphere alone, potential at the center relative to infinity should be kq/R (positive because I'm going against electric field?).

    With the dielectric in place to still take zero at infinity I need to take into account the potential difference across the dielectric (I think?)

    Edielectric = q/KAε0

    V(R->2R) = -∫Edielectric*dr (negative because I'm going in direction of electric field?) = -kq/Kr [2R - R] = -kq/2KR + kq/KR = kq/2KR

    I know that a dielectric will decrease the electric potential, so I thought V = Vsphere - Vdielectric = kq*((1/R) - (1/6R)).

    This is not correct and I'm not sure where I messed up.
     
  2. jcsd
  3. Feb 2, 2017 #2

    haruspex

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    I agree with your work up to this point:
    Think through that again. I believe you have made quite a simple mistake in what to subtract from what.
     
  4. Feb 3, 2017 #3

    rude man

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    Maybe more straightforward if you just integrated the E field from infinity to R.
     
  5. Feb 12, 2017 #4
    So integral of E field from infinity to R = (integral of E field from infinity to 2R) + (integral of E field through dielectric).

    From from infinity to 2R, can I still consider the system as a point particle and say the electric field is kQ/R^2? Or does that change with the dielectric in place?
     
  6. Feb 12, 2017 #5

    TSny

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    Answer this by applying Gauss' law to a Gaussian surface of radius r>2R.
     
  7. Feb 12, 2017 #6
    EA = Qenclosed/E0

    If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.

    E = Q/AE0 = kQ/R^2
     
  8. Feb 12, 2017 #7

    haruspex

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    Yes, but I'm not sure why this thread has become so complicated.
    A dielectric diminishes the potential difference across it. It exerts no long range field, so cannot affect potential gradients outside itself, as you confirmed by following TSny's hint.
    In post #1 you found the dielectric layer reduced the potential difference across it from kq/2R to kq/2KR. What reduction in potential in the sphere is needed to keep the potential at infinity zero?
     
  9. Feb 12, 2017 #8

    rude man

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    good thought!
    but don't say "E0" when you mean ε0.
     
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