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Simple Parallel-Plate Capacitor Question please HeLP

  1. Jan 12, 2008 #1
    Simple Parallel-Plate Capacitor Question...please HeLP!!!

    1. The problem statement, all variables and given/known data
    A parallel-plate capacitor is formed from two 2.32 cm × 2.32 cm electrodes spaced 2.85 mm apart. The electric field strength inside the capacitor is 85600000 N/C. What is the charge (in nC) on each electrode?

    2. Relevant equations

    3. The attempt at a solution
    E = Q / Eo * A
    A = 0.0232 * 0.0232 = 5.38e^-4
    Q = (E * Eo * A) 8 1,000,000,000 = 407.75 nC
    What am I doing wrong?
  2. jcsd
  3. Jan 13, 2008 #2
    Do I divided 407.75 by 2? Since there are 2 electrodes? I don't know what else I could be doing wrong.
  4. Jan 13, 2008 #3
    Thats not right either...does anyone know where I am going wrong???
  5. Jan 13, 2008 #4


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    Homework Helper

    you multiplied incorrectly, I guess. plugging into my calculator, I get 407.942041 nC

    or, to the right number of sig figs 408 nC
  6. Jan 13, 2008 #5
    How did you get that?
    K = 9 x10^9
    Eo = 8.85 x10^-12
    A = .0232 *.0232
  7. Jan 13, 2008 #6


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    Homework Helper

    K? That doesn't come in anywhere... I used E and A and epsilon_0. Just multiply them all together.
  8. Jan 13, 2008 #7
    I know I put that to see if you were using the same number to get Eo...I am not getting the right answer though
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