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Electic field / capacitance of parallel plates

  1. Sep 22, 2010 #1
    Upon calculating the electric field for a plate of charge, you arrive at:

    E = Q/Ae (Q the charge on the plate, A the area of the plate, e the permittivity of the dielectic)

    But then, isn't the total electric field between the plates twice this because of the equal but opposite charge on the other plate?

    I always see it written without a factor of 2.
  2. jcsd
  3. Sep 23, 2010 #2
    which way is the E field going , make a Gauss pillbox around the capacitor .
  4. Sep 23, 2010 #3


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    Is that equation only for the charge on 1 plate, or for both plates? Looks to me like it is only for the 1 plate based on your post.
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