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Homework Help: Two Protons Move From Location A and B to C

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Two protons are released from rest, one from location 1 and another from location 2. When these two protons reach location 3, the first proton has a speed that is 4 times the speed of the second proton. If the electric potentials at locations 1 and 2 are 231 V and 115 V, respectively, what is the electric potential at location 3?

    2. Relevant equations
    V =kq/r

    3. The attempt at a solution
    I honestly don't know where to start. I feel like I have way too few pieces of information.
  2. jcsd
  3. Dec 8, 2013 #2


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    Use the equation [itex] \Delta E_{kinetic}=q\Delta V [/itex].
  4. Dec 9, 2013 #3
    Oh that's a nice equation to have.
  5. Dec 9, 2013 #4
    Wait, with that equation, I can solve for ΔE, but what can I do with that?
  6. Dec 10, 2013 #5


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    You know that the speed of one of the protons is 4 times the speed of the other,so you can find out that its kinetic energy is how many times the other one's.And from this and the equation I gave you,I think you can find your answer.
  7. Dec 10, 2013 #6
    Alright so I now have ΔE. This is equal to k\frac{Q}{r2} . Is there anything else it's equal to? Because with that equation, all I can do is calculate r and that has nothing to do with kinetic energy.
  8. Dec 10, 2013 #7
    P.S. - What's wrong with my LaTeX code?
  9. Dec 10, 2013 #8
    I seriously have no idea what to do. I've been just staring at this problem. I don't have any r values and I can't calculate field strength or voltage without those. I don't have anything I need except the charge of a proton and the mass of a proton. That's it. What am I missing?
  10. Dec 10, 2013 #9
    You don't need any specific formula for electric field or voltage.
    All you need is the work energy formula as given by Shyan.
    Write it down for each proton. The initial KE is zero for both protons. And write ΔV as a difference, V3-V1 for example.
  11. Dec 10, 2013 #10
    Wait is ΔE kinetic energy in the formula given?
  12. Dec 10, 2013 #11
    Assuming that is the case, this problem and the help Shyan has been giving me makes much more sense. I now have:
    4KE2 = 1.6E-19(V3 - 231)
    KE2 = 1.6E-19(V3 - 115)

    Now I can solve for V3 no problem.
  13. Dec 10, 2013 #12
    Wrong answer. Calculated as such:
    1.6E-19(V3 - 231) = 4(1.6E-19(V3 - 115))
    V3 = 76.33

    That's wrong, where did I fall of the bus?
  14. Dec 10, 2013 #13


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    [itex] KE_1=\frac{1}{2}mv_1^2 [/itex] so if we have [itex]v_2=av_1[/itex] then,[itex] KE_2=\frac{1}{2}mv_2^2=\frac{a^2}{2}mv_1^2=a^2 KE_1[/itex] so if you multiply the speed by 4,kinetic energy will be multiplied by 16!
  15. Dec 10, 2013 #14
    I've never seen someone use an a in this equation. What does it represent?
  16. Dec 10, 2013 #15
    1.6e-19(V3 - 231) = 16(1.6e-19(V3 - 115))
    V3 = 12251.27 V

    Also incorrect.
  17. Dec 10, 2013 #16


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    a was just a representative for any coefficient,like your 4!
  18. Dec 10, 2013 #17
    Okay thanks. My WebAssign has a "Practice Another Version" and within it a "Show Solution". I always come to you guys because you let me figure it out, but this time it doesn't seem to be working. I'm just too incompetent on this topic. So anyway, here's what the "Show Solution" says:

    "Let the electric potential at location 3 be V3. From conservation of energy we know that
    ΔPE + ΔKE = 0
    where KE is the kinetic energy of the proton and PE is the electric potential energy. We also know that
    ΔPE = qΔV.

    We combine the above two equations to get
    qΔV + ΔKE = 0
    q(Vf − Vi) = −(KEf − KEi).
    Since the protons are released from rest, we have
    KEi = 0.
    We are given that
    v1 = 3v2
    and therefore
    KE1f = 9KE2f.

    We now apply the conservation of energy principle to each proton keeping in mind that proton 1 moves from location 1 to location 3 while proton 2 moves from location 2 to location 3.

    Proton 1:
    q(V3 − V1) = −KE1f

    Proton 2:
    q(V3 − V2) = −KE2f

    Dividing one equation by the other we get
    q(V3 − V1)
    q(V3 − V2)
    Substitute the values
    V1 = 231 V, V2 = 115 V, and KE1f = 9KE2f
    and solve for V3.
    V3 = 101 V
    This makes sense, as a positive charge will gain kinetic energy as it moves from a high potential point to a low potential point."

    Why are they dividing the entire equations?
  19. Dec 10, 2013 #18
    Dividing the equations is one way to proceed with solving them. But you don't need to do it.

    How did you do it yourself?
    Sow how you got that number in your previous post. The equation was OK. Probably you had some mistake in the algebra.
  20. Dec 10, 2013 #19
    All right. Excuse the faux scientific notation, it's just easier and avoids other sources of error.

    1.6e-19(V3 - 231) = 16(1.6e-19(V3 - 115))

    1.6E-19V3 - 3.696E-17 = 25.6E-19V3 - 2944E-17

    2.94E-14 = 24E-19V3

    V3 = 12251.27 V
  21. Dec 10, 2013 #20
    I think that the error is in the term 2944e-17

    But it is a really bad idea NOT to simplify the value of q right in the beginning. It will be less likely to make mistakes if the numbers are nicer.
    But the method is OK.
  22. Dec 10, 2013 #21
    What do you mean "simplify"? It's already only two sig figs.
  23. Dec 10, 2013 #22
    Correct answer!
  24. Dec 10, 2013 #23
    I mean that when you have
    you divide both sides by q. No need to carry this through all your calculations.
    Then you have

    What do you mean by "Correct answer"?
  25. Dec 11, 2013 #24
    Oh okay. I've shied away from doing that because there were a couple situations in which that didn't work for some reason. I meant that by fixing the 2944E-17, I had gotten the correct answer.
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