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Homework Help: Oil drop and electric field

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A pair of horizontal parallel plates are placed a small distance, 4.50 mm, apart in air. Each plate is rectangular with a width of 10.0 cm, and length of 46.0 cm. The potential on the upper plate relative to the lower plate is 1.03 × 103 V.

    A tiny drop of oil with an excess charge of –1.50 × 10–17 C and mass of 4.00 × 10−15 kg is placed just above the lower plate. The oil-drop is then released so that it is free to move under the action of the electric field.

    What time elapses in its transit from the lower to upper plate?

    • the magnitude of the electrical force it experiences is 3.43-12 N.
    • the magnitude of the electric field between the plates is 229000 v/m.
    • the magnitude of the resultant acceleration of the charge is 858 m/s2
    electronic charge = 1.60 × 10−19 C.

    P.S. neglect all non-electric effects such as gravity and air resistance.

    3. The attempt at a solution

    [tex]v_f=\sqrt{2ad}= \sqrt{2 \times 858 \times 0.46} =28[/tex]


    But the correct answer must be 0.00324 s, why? I can't think of any other way of solving this problem. I appreciate any help!
  2. jcsd
  3. Aug 22, 2010 #2
    1) The oil drop needs to travel 4.5mm, not 46cm.

    2) Your formula for t would only be true if v was the average velocity, which is not what you calculated. Use the formula s (or d) = at^2 / 2.

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