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Electric Potential Energy of protons

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Two protons that are very far apart are hurled straight at each other, each with an initial kinetic energy of 0.16 MeV, where 1 mega electron volt is equal to 1x10^6 multiply (1.6x10^-19) joules. What is the separation of the protons from each other when they momentarily come to a stop?


    2. Relevant equations

    Ue = (1/4[tex]\pi[/tex][tex]\epsilon[/tex])((q[tex]_{1}[/tex]q[tex]_{2}[/tex])/r)
    E[tex]_{f}[/tex] = E[tex]_{i}[/tex] + W

    3. The attempt at a solution

    I am very confused on how to get this problem started, especially since they give the kinetic energy. I tried to manipulate the kinetic energy formula ((1/2)mv^2) but I am still really confused on how to approach this problem.
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2

    kdv

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    Conservation of energy

    Total Kinetic energy at the beginning + potential energy at the beginning = total kinetic energy at the final position plus potential energy at the final position

    The initial position is when they each have the kinetic energy you gave but they are very far from each other. The final position is when they are at the point of closest approach
     
  4. Mar 10, 2008 #3
    Ohh alright so do I still do,

    e = kq^2/r where r is distance between protons
    k is 9*10^9
    0.16 * 1*10^6*1.6*10^-19 = 9*10^9 * (1.6*10^-19)^2 / r
    then I got r as (9x10^-15)

    the thing is that I got it wrong when i did it this way.
     
  5. Mar 10, 2008 #4

    kdv

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    Each has an energy of 0.16 MeV so your initial total kinetic energy is twice what you wrote
     
  6. Mar 11, 2008 #5
    ok! I got the separation as 4.5x10^-15 m ! Thank you!
     
  7. Mar 11, 2008 #6

    kdv

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    You are very welcome!:smile:
     
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