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What creates polarisation field when -div(P) = 0?

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Context: Between the two conducting cylinders of a coaxial cables, there is a dielectric hollow cylinder. There is free static charge Q on a length L of the inner conductor, and free static charge -Q on a length L of the outer conductor. Upon calculating that the polarisation field P = K/(ρ) (where the constant K accounts for charge, permittivity etc.. and ρ is the variable distance from the center). We are then asked to find the bound volume charge density within the dielectric.

    2. Relevant equations

    -div (P) = ρb

    3. The attempt at a solution

    Upon calculating the divergence, I find ρb = 0 . This is confirmed by the solutions manual.
    But I don't understand how this can be possible. The P field lines aren't closed lines as far as I know (unlike B field lines). They source at bound negative charges and sink at bound positive charges. How can there be a P field if there is no polarisation charge density spawning it?
    Please tell me if any of the assumptions I make above are wrong. Also, I suspect this might have something to do with distinguishing between surface and volume polarisation charges, but I don't see how that changes things.
     
  2. jcsd
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