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Electric Feild and Capacitance

  1. Feb 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A homemade capacitor is assembled by placing two 9inch pie pants 5cm apart and connecting them to the opposite terminals of a 9V battery. Estimate A) the capacitance, B) the charge on each plate c) the electric field halfway between the plates d) the work done by the battery to charge the plates.


    2. Relevant equations
    C=Eo(A/d) V=Ed Q=VC W= Q(V/2) PE= 1/2(QV)= 1/2(CV^2)= 1/2(Q^2/C)


    3. The attempt at a solution
    I was able to use the first formula I listed to solve for part A finding C= 7 x 10^ -12 and the third formula to solve for part Q= 7x 10^-11 but I can't seem to get part c. I tried using V=Ed but it doesnt give me the answer I'm looking for which should be 200 V/m.
     
  2. jcsd
  3. Feb 17, 2008 #2

    Doc Al

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    Staff: Mentor

    What answer are you getting? (What if you round it off?)
     
  4. Feb 17, 2008 #3
    I was getting 360 but then I realized that you probably want to divide the 9/2 since the voltage will be split evenly between each plate so then you end up getting 180..which may be close enough to 200??
     
  5. Feb 17, 2008 #4

    Doc Al

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    Staff: Mentor

    Use the equation V = Ed to solve for E.
    That's what I'd say. 180 rounded to one significant figure is 200.
     
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