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Parallel-plate capacitor

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Air "breaks down" when the electric field strength reaches 3 x 10^6 N/C, causing a spark. A parallel-plate capacitor is made from two 4.0cm-diameter disks. How many electrons must be transferred from one disk to the other to create a spark between the disks?

    2. Relevant equations

    Q = [tex]\epsilon[/tex]0AE

    N = Q/e

    3. The attempt at a solution

    Ok, I think this is right but I am not sure:

    Q = [tex]\epsilon[/tex]0AE

    A=[tex]\pi[/tex]r^2

    =(8.85x10^-12 C^2/Nm^2)([tex]\pi[/tex](0.02m)^2)(3x10^6 N/C)

    =3.3x10^-8 C = 33nC

    N = Q/e

    = (3.3x10^-8 C)/(1.60x10^-19 C/electron)

    = 2.1x10^11 electrons
     
  2. jcsd
  3. Feb 12, 2009 #2
    I'm trying to figure out this same problem only with diff. numbers and I'm stuck.. so if anyone has any ideas, please share!
     
  4. Feb 12, 2009 #3

    LowlyPion

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    Homework Helper

    Perhaps if you post your problem, with what you are having trouble reconciling, someone may be able to help clarify things?
     
  5. Feb 12, 2009 #4
    The problem is already posted above.
     
  6. Feb 12, 2009 #5

    LowlyPion

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    OK.

    What don't you understand about the solution he posted?
     
  7. Feb 12, 2009 #6
    It's not correct, is it?
     
  8. Feb 12, 2009 #7

    LowlyPion

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    What step do you think is incorrect?
     
  9. Sep 23, 2009 #8
    I used the same process! The only thing i did differently was use more significant figures. When done in this manner you get 2.08335*10^11 electrons.

    The correct equations were used to solve this problem.
     
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