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Monster E&M Question (Coulomb's Law, Electric Potential, Kinematics)

  1. Oct 6, 2013 #1

    kosovo dave

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    1. The problem statement, all variables and given/known data
    A proton (charge +1e and mass 1.67e-27kg) and an alpha particle (charge +2e and mass 6.64e-27kg) are placed 3 fm (1 fm=10e-15m) apart.
    a)what is the force on each particle
    B)what is the potential energy of the system?
    c)what is the acceleration of each particle at the beginning?
    d)after a long time the two particles are far apart. what is the speed of each particle at this point?


    2. Relevant equations

    [itex]F = ( q_1 q_2) / 4 π ε_0 r^2 = (k q_1 q_2) / r^2[/itex]
    [itex]U = q v = ( q_1 q_2) / 4 π ε_0 r[/itex]
    [itex] K E = 1 / 2 m v^2[/itex]
    [itex] F = E q = m a [/itex]
    3. The attempt at a solution
    a) Using Coulomb's Law, I found the force to be 51.2 N.
    b) substituting my values into the second equation above, I got that the potential energy of the system is 1.5e-13 J.
    c)Using the fourth equation above I found the accelerations. For the proton I got 3e28 m/s^2. For the alpha particle I got 8e27 m/s^2.
    d) This part got a little tough. Realizing that the acceleration decreases as the particles move apart, I tried to integrate their acceleration from their initial separation out to infinity, but this quickly proved to be tedious/confusing. Looking at it again, I realized that as the particles move towards infinity all of the potential energy will be converted into kinetic energy. So I solved equation 3, substituting my answer from b) in for the kinetic energy and the mass of each particle. I found the velocity of the proton to be 1.34e7 m/s and the velocity for the alpha particle to be 6.72e6 m/s.

    Did I do this correctly? This question is from an old test my prof gave us to study (I don't have the solution) and I have no intuition for how quickly these types of particles should be moving.
     
  2. jcsd
  3. Oct 6, 2013 #2
    Yes.The procedure you did is correct.
     
  4. Oct 6, 2013 #3

    kosovo dave

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    Thanks! If I wanted to continue on my initial path of integrating (no pun intended) the acceleration to get the final velocity, could somebody show how that would work?
     
  5. Oct 6, 2013 #4
    What the supposed pun supposed to be?
     
  6. Oct 6, 2013 #5

    kosovo dave

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    Gold Member

    path of integrating = path integral. not a good one, but gimme a break -- I've been studying all day.
     
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