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Positively charged oil drop

  1. May 7, 2007 #1
    A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates. If the electric force on the drop is found to be 8.60 × 10−16 N and the electric field magnitude is 153 V/m, what is the magnitude of the charge on the drop in terms of the elementary charge e?

    The correct answer should be 35e, but I am not getting it.

    My attempt:
    I tried the equation
    F = Eq
    8.60 × 10−16 N = (153 V/m)(q)
    q = 5.6 x 10-18 N

    Then I used
    e = q/k
    e = 5.6 x 10-18 N / 8.99 x 10^9

    which does not give me the right answer
  2. jcsd
  3. May 7, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Good :smile:
    Not so good. I don't know where you got that equation from, but to express your charge in terms of elementary charges, you have to divide through by the value of the elementary charge (1.6x10-19 C).
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