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Potential Difference in particle

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle with a charge of -1.5 µC and a mass of 3.0 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 50 m/s.

    (a) What is the potential difference VB - VA between A and B?

    2. Relevant equations


    v=(square root of) -2Qo(Va-Vb)/m
    3. The attempt at a solution
    everything else is given in the equation, v=50m/s, Q=-1.5 x 10^-6 C, m=3 x 10^-6, but i cannot figure out how to get Va-Vb by itself. I tried squaring both sides which gave me Va2=-4Qo2(Va-Vb)2/m, but from there i'm stuck
     
  2. jcsd
  3. Jul 11, 2008 #2

    kreil

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    Isn't it just algebra?

    [tex] v= \sqrt{ \frac{-2Q_0 (V_a - V_b)}{m}}[/tex] (square both sides)

    [tex] v^2 = \frac{-2Q_0 (V_A - V_B)}{m}[/tex](multiply/divide to move things around)

    [tex] -\frac{ v^2 m}{2 Q_0}= V_A-V_B[/tex]
     
  4. Jul 11, 2008 #3

    Redbelly98

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    You might try thinking about conservation of energy (that is where the equation you give comes from anyway):

    (Change in kinetic energy) = - (Change in electric potential energy)
     
  5. Jul 11, 2008 #4
    oh ok, i'll try that
     
  6. Jul 12, 2008 #5
    i plugged in the numbers and got -3 x 10^-6, but that was not right.
     
  7. Jul 12, 2008 #6

    Redbelly98

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    I plugged in the numbers (Kreil's last equation in msg #2) and got something else. Try again, be careful about properly entering the numbers into your calculator.

    Hint: the answer is a lot larger than 1V.
     
  8. Jul 13, 2008 #7
    ohh, a mental error on my part. got it, thanks!
     
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