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Homework Help: Calculate the smallest possible distance between the two electrons

  1. Nov 11, 2007 #1
    Question:
    Two electrons are fired at 3.5 x 10^6 m/s directly at each other.
    (a) Calculate the smallest possible distance between the two electrons.

    Attempt:
    I tried Ee=Ek for this question and this is how it went.
    Note*(q1=q2)

    Ee=Ek
    kq^2 = 0.5mv^2
    r
    r= kq^2
    0.5mv^2
    = (9.0x10^9(N x m^2)/C^2)(1.6x10^-19C)^2
    0.5(9.1x10^-31kg)(3.5x10^6m/s)^2
    = 4.1x10^-11m


    I don't seem to be getting the right answer. Can someone please tell me what I'm doing wrong?
     
  2. jcsd
  3. Nov 11, 2007 #2

    hage567

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    Homework Helper

    Your initial kinetic energy is not just 0.5mv^2, it's double that, since both electrons are moving. Try that.
     
  4. Nov 11, 2007 #3
    The answer still doesn't match, but I'm thinking the book is wrong. So thank you=)
     
  5. Nov 11, 2007 #4
    what is the answer?
     
  6. Nov 11, 2007 #5
    It says 4.5x10^-6m.
     
  7. Nov 11, 2007 #6
    i did it quickly, and i got the same answer u got in the first part. I did it another completely different way. interesting.. well i gtg.. i will try to look at this problem later.. but i dont think its right to assume the book is wrong.
     
  8. Nov 11, 2007 #7

    hage567

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    Homework Helper

    They've taken the square root. I don't know why.
     
  9. Nov 11, 2007 #8
    That would work if the electrostatic force was used, but I can't see how that fits.
     
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