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Homework Help: Free surface charge density (not bound density)

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data
    a parallel plate capacitor has 2 plates separated by a dielectric of rel. permittivity 5.0 are separated by 0.20mm and have area of 5.0 cm2.
    Potential difference between the plates is 500V.

    I need to be able to calculate the free surface charge density.

    2. Relevant equations
    I'm assuming Dout = Din

    From boundary conditions D1 ┴ = D2
    and D2 ┴ - D1 ┴ = σf

    Could someone just confirm what the boundary conditions are so I can say that the free surface charge density is zero or not!

    3. The attempt at a solution

    I have Dout = Din = ε0Ein + P
    so from calculations
    P = 4.43x10-9 - (8.85x10-12 x 100)
    P = 3.54x10-9 Cm-2

    Could I just say;
    D1 ┴ = D2 ┴ so there is zero free surface charge density!
  2. jcsd
  3. Sep 20, 2014 #2
    Sorry, I should have said.
    D1 ┴ is the electric displacement in media 1 perpendicular to a boundary with a different media
    D2 ┴ is the electric displacement in media 1 perpendicular to a boundary with a different media

    They form the boundary condition
    could have been
    D1 = D2.

    and P is the polarization of the material.
  4. Sep 21, 2014 #3

    rude man

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    Homework Helper
    Gold Member

    I don't know what you mean by "different media" or "Dout" vs. "Din". There is only one medium, the dielectric of epsilon_relative = 5.

    D is continuous from one medium to another along the plates' normal even if there were two or more media, but there aren't.

    So V = E x d, d = 0.2mm so you know E.
    Then, D = epsilon x E.
    And if you know D you know the surface charge density, right?

    This ignores fringing effects of course which you have to do since they didn't give you the dimensions of the plates.
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