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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?

    Given:
    • 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]

    [tex]t=\frac{l-0}{v}=\frac{0.46}{28}=0.01642[/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.

    -Tusike
     
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