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Magnitude of charge on oil drop (electric field)

  1. Mar 21, 2007 #1
    A positively charged oil drop of mass 1.00 × 10−15 kg is placed in the region of a uniform electric field between two oppositely charged, horizontal plates. The drop is found to remain stationary under the influence of the Earth's gravitational field and the uniform electric field of 0.680 × 104 N/C. What is the magnitude of the charge on the drop?

    I am not sure where to start. I am given a mass so would I start with

    U = -(G(m1)(m2))/r

    Thanks in advance
  2. jcsd
  3. Mar 21, 2007 #2
    If the drop is stationary, what can you say about the net force on the drop?
  4. Mar 21, 2007 #3
    would force = 0?
  5. Mar 21, 2007 #4
    do you mean net force?
  6. Mar 21, 2007 #5
    Sorry, I'm still confused.
    Wouldn't gravity be the force that allows it to drop?
    Which equation would be good to use?
  7. Mar 21, 2007 #6
    The force of gravity acts down on the drop, but the drop is charged and it sits (stationary) between two charged plates. Since it is stationary, what can you say about the acceleration and net force on it?
  8. Mar 21, 2007 #7
    There is no acceleration and thus net force = 0
  9. Mar 21, 2007 #8
    Good, now what two forces act on the drop?
  10. Mar 21, 2007 #9
    gravity and drag force?
  11. Mar 21, 2007 #10
    Drag force??
    A positive charge is in an electric field.
  12. Mar 21, 2007 #11
    gravity and the weight of the drop
  13. Mar 21, 2007 #12
    those are the same character!(force)
    I gave you a big hint.
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