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Parallel plates capacitance

  1. Jul 19, 2004 #1
    A parallel plate capacitor has charge Q and plates of area A. What force acts on one plate to attract it toward the other plate?

    It's F = Q^2 / (2*8.854e-12*A) ... something to do with the electric field divided in 2? Wondering why there's a 2 in the denominator... thanks :)
     
    Last edited: Jul 19, 2004
  2. jcsd
  3. Jul 20, 2004 #2

    Doc Al

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    Staff: Mentor

    To find the force that one plate exerts on the other, first find the field created by the charge on that plate. That's where the 2 will come in.
     
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