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Protons released from rest in a vacuum.

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    If two protons start out at rest relative to each other, separated by 0.1 nanometers in vacuum, and then they are released, calculate their relative positions after 10 seconds.

    2. Relevant equations

    F = (kq^2)/r^2 = m*a

    3. The attempt at a solution

    I am fairly certain this needs to be solved using an integral since the force is constantly changing with the distance between the two protons, but I'm quite confused what it would have respect to. Time perhaps, ranging from 0 to 10 seconds and using a = dv/dt.
  2. jcsd
  3. Jan 19, 2010 #2


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    First, we are assuming that the protons can be described classically, correct?

    Second, what do we know about position (in this case r) and its relation to acceleration (a)?
  4. Jan 19, 2010 #3
    Yes assuming a classically described proton. What is known about position in relation to acceleration is at time t = zero seconds, acceleration is equal to zero. At the instant the protons are released acceleration would be F/m which would be (k*q^2)/(m*r^2) where k = 8.89*10^9, q = 1.602*10^-19, m = 1.673*10^-27, and r = .1nm so a = 8.513*10^27 but as the protons become further apart the value of r increases, reducing F and therefor reduces the acceleration.
  5. Jan 19, 2010 #4


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    What I was trying to get at is dr/dt=v and dv/dt=a

    Using initial conditions (r(0)=.1nm, v(0)=0, a(0)=kq^2/m(.1nm)^2), and after you have established the correct and useful coordinates, you should get a function of r in terms of a, or vice versa.
  6. Jan 19, 2010 #5
    So using dr/dt=v and dv/dt=a and a=kq^2/mr^2 you could integrate a with respect to t twice to find v and then r in terms of a time t?
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