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Charge in a capacitor

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Two parallel plates are placed 0.10 m apart with one vertically above the other and their edges aligned. The potential difference of the upper plate is 100 kV relative to the lower plate. What charge must a spherical raindrop of diameter 1.0 mm carry if it remains suspended between the plates? Assume that the electric field between the plates is uniform and give your answer in units of the charge on an electron. (The density of water is 1000 kg m−3

    2. Relevant equations
    C = q/V
    C = e0 et(A/D)

    3. The attempt at a solution
    I know that I've got to work out the capacitance in order to work out the charge q=CV but how do I find the capacitance?

    I understand Capacitance = permitivity of free space x relative permitivity (Area/Distance of plate separation)

    I think I need to find the area and the permitivity but I dont understand why they've given the density of water?

    Help please!!!
     
  2. jcsd
  3. Jan 30, 2012 #2

    tiny-tim

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    hi shyguy79! :smile:

    i think you're making this too complicated :redface:

    forget about capacitance, it's just a mechanics question …

    what is the definition of voltage? :wink:
     
  4. Jan 30, 2012 #3

    ehild

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    You do not need the charge on the capacitor.
    There is a raindrop suspended between the plates. It carries so much charge that the resultant force on it is zero. The charge of the raindrop is the question.

    ehild
     
  5. Jan 30, 2012 #4
    Err... The 'work' done to move a charge between to points? Why?
     
  6. Jan 30, 2012 #5

    tiny-tim

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    ok! :smile:

    now balance that against the work done by gravity :wink:
     
  7. Jan 30, 2012 #6
    So why are we given the density of the water, the separation of the plates, the diameter of the raindrop and the potential between the plates? This is so frustrating!

    Thank you all for your help!
     
  8. Jan 30, 2012 #7
    So to balance the raindrop then the charge x electric field = mass x gravity? Or something!!!
     
  9. Jan 30, 2012 #8

    tiny-tim

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  10. Jan 30, 2012 #9
    But the mass of the raindrop? And why the distance between the plates
     
    Last edited: Jan 30, 2012
  11. Jan 30, 2012 #10

    tiny-tim

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    volume times density? :wink:
     
  12. Jan 30, 2012 #11
    Ahhhhhhh!!!!! Gotcha!!
     
  13. Jan 30, 2012 #12
    And so the Electric field = - Δv / Δx ?
     
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