Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?