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Three protons and two positrons...

  1. Feb 23, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    There protons are and two positrons are held such that two protons are on the ends of a diagonal of a square and the two positrons are on the ends of the second diagonal. The third proton is situated at the centre of the square. The system is released from rest. Find the kinetic energy of the three protons after a long. (Side of square is ##a##)

    2. Relevant equations
    $$F=\frac{ke^2}{a^2}$$

    3. The attempt at a solution
    I tried solving the problem by assuming that final potential energy is ##0## since all particles except the proton at the centre of square will move to infinity due to rulsions and initial potential energy is $$\frac{4ke^2}{a}+\frac{2ke^2}{a\sqrt{2}}+\frac{4ke^2}{\frac{a}{\sqrt{2}}}$$

    But in solution given, they have state that only the positrons will fly away and the protons will remain. How is this possible?
     
  2. jcsd
  3. Feb 23, 2016 #2

    TSny

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    Does the solution actually say that none of the protons gain any KE, even after a long time?
     
  4. Feb 23, 2016 #3

    Titan97

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    No. Here is the picture:

    photo-1.JPG
     
  5. Feb 23, 2016 #4

    TSny

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    OK. Compare the mass of a proton to the mass of a positron.
     
  6. Feb 24, 2016 #5

    Titan97

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    The mass of the positron will not affect the force with which it will repel the proton.
     
  7. Feb 24, 2016 #6

    TSny

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    Why not?
     
  8. Feb 25, 2016 #7

    Titan97

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    Because F=kq/r^2.
     
  9. Feb 25, 2016 #8

    TSny

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    OK.

    At the instant of release, how does the magnitude of the net force on a positron compare to the magnitude of the net force on a proton?

    At the instant of release, how does the magnitude of the acceleration of a positron compare to the magnitude of the acceleration of a proton?
     
  10. Feb 27, 2016 #9

    Titan97

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    I get it now. Acceleration of positron is far greater than that of the proton.
     
  11. Feb 27, 2016 #10

    TSny

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    Yes. To a good approximation, you can neglect the motion of the corner protons during the time that the positrons move far enough away to have little influence on the protons.
     
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