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Homework Help: Capacitor filled with 2 different dielectrics

  1. May 19, 2010 #1


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    1. The problem statement, all variables and given/known data
    Hi there, I got a little problem with this one:
    plate capacitor is filled with 2 different dielectrics with permittivities [tex]\epsilon_1, \epsilon_2[/tex] (surrounding environment is [tex] \epsilon_0 [/tex]) and their respective thicknesses are [tex] d_1 [/tex] and [tex] d_2 [/tex]. The layers are parallel to the plates of the capacitor. The voltage between the electrodes is U and their surface is S. What are the forces acting upon plates ?

    2. Relevant equations
    gauss' law for electric displacement field; F = qE; U=d E (d = distance between plates)

    3. The attempt at a solution
    I tried this:
    electric displacement field is the same in whole capacitor [tex] D = \sigma [/tex] where [tex] \sigma [/tex] is surface density of charge on plates.
    The voltage is then [tex] U = \frac{D}{\epsilon_1}d_1 + \frac{D}{\epsilon_2}d_2 [/tex].

    From the last equation, using the 1st one, the charge density is:
    [tex] \sigma = U \frac{\epsilon_1 \epsilon_2}{d_1 \epsilon_2 + d_2 \epsilon_1} [/tex].

    And the force on plate at dielectric 1: [tex] F = S \sigma E_0 = \frac{S}{\epsilon_0} \sigma^2 = \frac{S}{\epsilon_0}U^2 (\frac{\epsilon_1 \epsilon_2}{d_1 \epsilon_2 + d_2 \epsilon_1})^2 [/tex]
    [tex]E_0[/tex] is the field outside the dielectrics, since the plate is not in the dielectric.

    Which, according to the book, is wrong. The correct answer should be
    [tex] F = \frac{\epsilon_0^2}{2 \epsilon_1} (\frac{U}{d_1 + d_2})^2 S [/tex].

    I'm completely lost. The book gives no detailed explanation, and after 2 hours of scratching my head I need help. Thanks in advance for any suggestions.

    [edit: wrong names for electric field and electric displacement field... sorry for my english ;)]
    Last edited: May 19, 2010
  2. jcsd
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