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Identical Charges

  1. May 17, 2005 #1
    First of all I've read the sticky and I am not trying to get somebody to do my homework. I have been working on this problem for the last two weeks and have the piles of paper to prove it. Here is my problem: Two identical 7.5u-c charges start from rest and are initially spaced 5.5 cm from each other. If we fix the one charge, and let the other go free, how fast will it be moving when the charges are 30 cm apart? Assume the charges have mass 10^-6 kg. The solution is 3877 m/s.

    Roads I've been down:
    v^2= vay^2 + 2ay(y-y0)
    = 0 + 2ay(24.5)
    v= sqrt(2a(24.5))
    = 2632 m/s

    mv^2= KQ^2/r
    v^2 = KQ^2/rm
    v=sqrt((9*10^9)(7.5 X 10^-6)^2)/(0.055m)(1*10^-6))
     
  2. jcsd
  3. May 17, 2005 #2

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    What is the change in potential energy from the first point to the second (it goes down)? All of this is transformed into kinetic energy.
     
  4. May 17, 2005 #3
    I've been able to solve this problem for a large distance between the two charges.. in that case the potential difference is zero so I was able to get this:

    deltaK + deltaU = 0
    1/2mv^2 + 1/2mv^2 - 0 + Q(0-V) = 0 or
    mv^2 = Q(kQ/r) = kQ^2/r
    (1.0 * 10^-6kg)v^2 = (9 * 10^9)*(7.5*10^-6)/(0.055m)
    v = 3.0 * 10^3 m/s

    I'm not sure what the potential difference is at this distance.
     
  5. May 17, 2005 #4

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    It's just V1-V2, or kQ^2/r1 - kQ^2/r2.
     
  6. May 17, 2005 #5
    I apologize for my daftness but that does not yield the velocity when they are 30 cm apart.
     
  7. May 17, 2005 #6

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    Is your equation:

    1/2 mv^2 = kq^2 (1/(5.5cm) - 1/(30 cm)) ?

    If so, check your math, because that's right.
     
  8. May 17, 2005 #7
    Oh Thank You! Brain malfunction .. too long looking at the same problem.
    Very much appreciated! I can't believe the incredibly quick response. I will highly reccomend this site. All the best!
     
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