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Total kinetic energy of two protons

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Two protons are held fixed at a distance of 3 angstroms (3 x 10-10 m) from
    one another. The protons have a charge of +1.6 x 10-19 C. After they are released they
    repel each other and fly apart and each acquires some kinetic energy. What is the final
    total kinetic energy of the two particles when they are at a large distance from each other?

    2. Relevant equations

    Potential Electric Energy= (1/4pie)*q1q2/r
    Ki+Ui=Kf+Uf

    3. The attempt at a solution

    3A * (3e-10m/1 A) = 9e-10m
    Ue=(9e9 Nm^2/C^2)(1.6e-19C)(1.6e-19C)/(9e-10m)= 2.56e-19 Nm

    And I know that as r goes to infinity, potential energy goes to 0
    thus
    0+Ui=Kf+0
    2.56e-19J=Kf
    But the answer is 7.7e-19, what i am doing wrong, what step am I missing thank you!
     
  2. jcsd
  3. Oct 12, 2009 #2

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Welcome to PF :smile:

    When you multiply by (3e-10m/1 A) to convert units, that is saying that
    1 A = 3e-10m​
    and this is incorrect.

    Actually, you do not need to calculate the distance in m; the problem statement tells you that the protons are ____m apart.

    Hope that helps.
     
  4. Oct 12, 2009 #3
    Yes thank you very much! Wow that was really stupid of me
     
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