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Electrical Potential Force

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1350 km/s, measured relative to the earth.

    Find the maximum electrical force that these protons will exert on each other.

    2. Relevant equations

    Mass of Proton [tex] 1.67 * 10 ^{-27} [/tex]

    [tex] KE_{i} = PE_{f} [/tex]

    [tex] \frac {1} {2} * m_{p}*v_{1}^{2} = \frac {1} {4*pi*8.85*10^{-12}}*\frac {Q_{2}} {R} [/tex]


    3. The attempt at a solution

    I'm not sure where to go from here.
     
  2. jcsd
  3. Feb 25, 2007 #2

    Doc Al

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    Find the distance of closest approach by setting the initial KE (total KE) equal to electric potential energy. Then use Coulomb's law to find the force.
     
  4. Feb 25, 2007 #3
    What is V_i ?
     
  5. Feb 25, 2007 #4

    Doc Al

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    What do you mean by V_i? The initial speed? That's given. The initial potential? Assume they start out infinitely far away from each other (for all practical purposes).
     
  6. Feb 25, 2007 #5
    To find the closet approach, I'm solving for R as you said so it should be in the form of:

    [tex] \frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}} = R [/tex]
     
  7. Feb 25, 2007 #6

    Doc Al

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    Looks to me like you are only counting the KE of one proton--use the total KE of both.
     
  8. Feb 25, 2007 #7
    When we count both, would it just be KE_1 + KE_2?

    [tex] 2*(\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}}) = R [/tex]
     
  9. Feb 25, 2007 #8

    Doc Al

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    That's right--just add up the KE. But fix that equation--I just realized that you have some factors inverted. The magnitude of the PE is kq^2/R.
     
  10. Feb 25, 2007 #9
    Ah lets show each step:


    [tex] KE_{i}1 + KE_{i}2 = PE_{f} [/tex]

    [tex] 2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R} [/tex]

    [tex] 2(m_{p}*v_{i}) = \frac {KQ^{2}} {R} [/tex]

    We divide out KQ^2 to isolate R

    [tex] \frac {2(m_{p}*v_{i})} {KQ^{2}} = {R} [/tex]

    ??
     
  11. Feb 25, 2007 #10

    Doc Al

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    Good.

    Typos in the right hand term.

    Almost--that should be v_i^2. And the 2 cancels.

    Do this step over. Note that R starts out in the denominator.
     
    Last edited: Feb 25, 2007
  12. Feb 25, 2007 #11
    [tex] m_{p}*v_{i} = \frac {KQ^{2}} {R} [/tex]

    Divide by [tex] m_{p}*v_{i} [/tex]

    [tex] \frac {KQ^{2}} {(m_{p}*v_{i})*R} [/tex]

    Multiply by R

    [tex] R = \frac {KQ^{2}} {(m_{p}*v_{i})} [/tex]

    Sorry my algebra is a little rusty
     
  13. Feb 25, 2007 #12

    Doc Al

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    Good.

    You mean:

    [tex]1 = \frac {KQ^{2}} {(m_{p}*v_{i})*R} [/tex]

    Good.

    No problem. Remember the cross-multiply trick. When you see this:

    [tex]\frac{A}{B} = \frac{C}{D}[/tex]

    You can convert to this:

    [tex]A*D = B*C[/tex]
     
  14. Feb 25, 2007 #13
    After I find the closet approach, you said to use Coulomb's Law to find the force. Is the R in here equilv to d in Coulomb's Law where:

    [tex] F = \frac {K*Q_{1}*Q_{2}} {d^{2}} [/tex]

    and F is what I'm after in t his problem?
     
  15. Feb 25, 2007 #14

    Doc Al

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    Yes and yes.
     
  16. Feb 25, 2007 #15
    [tex] R = \frac {KQ^{2}} {(m_{p}*v_{i}^{2})} [/tex]

    [tex] R = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(1.67*10^{-27})*(1350)^{2}} [/tex]

    [tex] R = \frac {2.30*10^{-28}} {4.50*10^{-24}} [/tex]

    [tex] R = 0.000051 [/tex]

    [tex] F = \frac {K*Q_{1}*Q_{2}} {d^{2}} [/tex]

    [tex] F = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(0.000051)^{2}} [/tex]

    [tex] F = 8.84 * 10^{-20} [/tex]

    Can someone verify my calculations, the online program says it's incorrect. I think I did everything right, I'm wondering if km/s needs to be converted into m/s?
     
    Last edited: Feb 25, 2007
  17. Feb 26, 2007 #16

    Doc Al

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    Absolutely.
     
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