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Finding the force exerted by one plate on another

  1. Oct 9, 2005 #1
    Hello everyone, I'm stuck on a mutliple choice question: A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 10-6 C then the force exerted by one plate on the other has a magnitude of about:
    1 E4 N
    9 E5 N
    5 N
    9 N
    0

    Here is what i did:
    I know the E-field right outside a conductor is E = (Q/A)/Eo <--permitivity. So i found E = (5e-6/.3)/8.85e-12 = 1883239 N/C; Then i knew F = QE, so I said, F = (5e-6)(1883239) = 9.4N, which was wrong, then i thought, hm..maybe i messed up a unit somehow, so i tried 9e5N, also wrong, then i thought, well if they are both positive, that means the net force is going to be 0, because ulike charges repel, same charge, 0 net force. also wrong. What did i do wrong?

    I also ran into odd problem, i'm wondering if the homework problem is wrong...
    A 20 F capacitor is charged to 200 V. Its stored energy is:
    4000 J
    0.1 J
    2000 J
    0.4 J
    4 J

    Easy enough, U = .5*CV^2;
    U = .5(20)(200)^2 = 400000J;
    The answer is .4J though?
     
    Last edited: Oct 9, 2005
  2. jcsd
  3. Oct 17, 2005 #2

    Tom Mattson

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    Staff Emeritus
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