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Dielectrics and guass' law

  1. Sep 18, 2007 #1
    the answer i keep getting is wrong ... help?

    1. The problem statement, all variables and given/known data

    Two parallel plates of area 110 cm2 are given charges of equal magnitudes 9.6 × 10-7 C but opposite signs. The electric field within the dielectric material filling the space between the plates is 4.2 × 106 V/m.

    (a) Calculate the dielectric constant of the material.
    (b) Determine the magnitude of the charge induced on each dielectric surface.


    2. Relevant equations

    k=q/(E*A*epsilon_0)

    q'=q(1-(1/k))

    3. The attempt at a solution

    k=q/(E*A*epsilon_0)
    =9.6E-7/(4.2E6*110E-4*8.85E-12)
    =2.3

    q'=q(1-(1/k))
    =9.6E-7(1-(1/2.3))
    =5.4E-7
     
  2. jcsd
  3. Sep 18, 2007 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    It is not clear but I suspect that these are non-conducting plates, in which case Gauss' law is:

    [tex]\oint E dA = E2A = \frac{Q}{k\epsilon_0}[/tex]

    AM
     
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