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Homework Help: Calculating the capacitance problem

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data
    If an uncharged parallel-plate capacitor (capacitance C) is connected to a battery, one plate becomes negatively charged as electrons move to the plate face (area A). The depth d from which the electrons come in the plate in a particular capacitor is plotted against a range of values for the potential difference V of the battery. When d is one picometre, the potential difference is 20V. The gradient is equal to 5x10-14. What is the ratio C/A.

    2. Relevant equations

    q=ε₀EA
    q=CV

    3. The attempt at a solution

    I combined the above two equations to get the expressions C/A=ε₀/D
    However that doesn't provide me with a numerical solution for the ratio C/A.

    Can anyone please help?
     
  2. jcsd
  3. Oct 18, 2008 #2

    Redbelly98

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    I don't understand what is meant by "the depth ... from which the electrons come in the plate".

    Also, what is this "gradient equal to 5x10-14" mean? And what are the units?
     
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