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Homework Help: Capacitance with Non-Uniform Dielectrics

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

    1. A parallel plate capacitor, with plates of area A, and spacing d, is filled with a non-uniform
    dielectric, with a permitivity that varies as

    ε = ε0 + ax

    where a is a constant, and x is the distance from one plate

    2. Relevant equations

    C = Q/V

    I'm assuming I'm allowed to assume this is a linear dielectric so

    P = XeE

    D = P + E = (1+Xe)E = εE

    Where P is the polarization vector and Xe is electric susceptibility and D is the electric displacement
    -grad(P) = ρbound

    P dot n = σbound
    Where n is a unit vector normal to the surface

    ε = ε0(1+χe)

    3. The attempt at a solution

    I am able to solve for Xe and sub it back into the equation for P


    P = ax/ε0E

    And I know that the E from the plates is σfree0

    And that C = Q/V and that the sum of the bound charges should be zero because otherwise conservation of charge would be violated so V should be changing.

    But I am really unsure what to do to figure out the total electric field.
    Last edited: Oct 13, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    Are you trying to solve for the capacitance or the E-field? I'm assuming the capacitance since that was one of your relevant equations and typically for a parallel plate capacitor, that's what you're looking for.

    A few equations to get you started:
    [itex]E_n*\epsilon(x) = \rho_s[/itex] where [itex]E_n[/itex] is the normal component of the E-field and [itex]\rho_s[/itex] is the surface charge density on the plates. If you use [itex]\rho_s[/itex] and substitute in for the charge and area, then solve for the potential V using:
    [itex]V = -\int_{0}^{d}E\cdot dl[/itex]
    You should be able to get the capacitance from there.
  4. Oct 17, 2011 #3
    I atually ended up figuring it out, but it was really crazily easy. Basically I was supposed to use the electric field from a parallel-plate capacitor and substitute in ε for ε0 then integrate to find the voltage.
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