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Homework Help: Paralle plate capacitor (with dielectric)

  1. Dec 6, 2006 #1
    EDIT: I figured it out.

    1. The problem statement, all variables and given/known data
    A parallel-plate capacitor of width [itex] 2w [/itex], length [itex] L [/itex] and separation [itex] d [/itex] is partially filled with dielectric material of [itex] \epsilon_0 \epsilon_r [/itex]. A voltage [itex] V_0 [/itex] is applied.

    1. Find [itex] \vec D [/itex], [itex] \vec E [/itex], [tex] \rho_S [/itex]

    ================ -------
    NOTE: The dielectric is on the right side (I colored it red)

    2. Relevant equations
    Note: Book convention uses [itex] \rho [/itex] as charge density.

    Boundary Conditions:
    [tex] B_{1t} = B_{2t} [/tex]
    [tex] D_{1n} - D_{2n} = \rho_S [/tex]

    [tex] \vec E = \frac{\vec D}{\epsilon} [/tex]

    Ideal Parallel Plate:
    [tex] \vec E = \hat e \frac{V_{12}}{d} [/tex]

    3. The attempt at a solution
    We first assume the use of ideal conditions, thus the expressions above hold.

    Next we note that the E field is directed from the top plate to the bottom plate. So we have,

    [tex] \vec E_1 = -\hat z \frac{V_{12}}{d} =-\hat z \frac{V_{0}}{d}[/tex]
    [tex] \vec E_2 = -\hat z \frac{V_{0}}{d} [/tex]

    The only component of the E fields are tangential components. ie,
    [tex] \vec E_1 = -\hat z E_t + \hat e E_n [/tex]
    Where: [itex] \hat e [/itex] is a basis vector in some direction, and [itex] E_n = 0[/itex]

    I'm basically trying to say that the E fields consist of no normal components to the interface.

    Thus we satisfy our first BC with [itex] \vec E_1 = \vec E_t [/itex].

    Moving on to the second BC we note:
    [tex] \vec D_1 = \epsilon \vec E_1 = \epsilon_0 \vec E_1 [/tex]
    [tex] \vec D_2 = \epsilon \vec E_2 = \epsilon_0 \epsilon_r \vec E_2 [/tex]

    [tex] \vec D_1 = \epsilon_0 \left(-\hat z \frac{V_{-}}{d} \right) [/tex]
    [tex] \vec D_2 = \epsilon_0 \epsilon_r \left(-\hat z \frac{V_{0}}{d} \right) [/tex]

    I'm confused here (and probably above too).

    [tex] \vec D_1 - \vec D_2 = -\hat z \frac{\epsilon_0 V_0}{d} (1-\epsilon_r) [/tex]
    Now the normal component is [itex] \hat -x [/itex] (is this right?). (The coordinate system I'm using is x is left to right, y is in and out of page, and z is up and down.)

    So doing the dot product:
    [tex] \hat n \cdot (\vec D_1 - \vec D_2) = \rho_S [/tex]
    This implies, [itex] \rho_S = 0 [/itex]

    ... I'm just confused here. I don't understand this part at all. Any help would be great!
    Last edited: Dec 6, 2006
  2. jcsd
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