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Electrical Potential Energy of three quark system

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data

    A proton is composed of three quarks: two "up" quarks, each having charge +2e/3, and one "down" quark, having charge -e/3. Suppose that the three quarks are equidistant from one another. Take the distance to be 3×10-15 m and calculate the potential energy of the subsystem of two "up" quarks. (MeV)

    At first I confused the formula below with [itex] \phi(P) = \Sigma^N_{i = 1} \frac{q_i}{r_i}[/itex] but this is the potential instead of potential energy at a point P with [itex]\phi[/itex]=0 taken at infinity?

    2. Relevant equations

    Total electrical potential energy

    [tex] U = \frac12 \Sigma_{i=1}^N \Sigma_{j \ne i}^N k \frac {q_i q_j}{ r_{i,j}}[/tex]

    3. The attempt at a solution

    For three charges in this arrangement I should just be able to do

    [tex] U = k\frac{q_1q_2}{r_{12}} + k \frac{q_1q_3}{r_{13}} + k\frac{q_2q_3}{r_{23}}[/tex]

    For the total potential energy should I just not be able to do this? It says this is incorrect but I'm not seeing how given that the total potential energy is just the summation of the potential energies individually and the potential between [itex]q_2[/itex] and [itex]q_3[/itex] should be the same as the potential between [itex]q_1[/itex] and [itex]q_3[/itex] because they have the same radii and charge proportions?

    [tex] U = [(8.99 * 10^9) \frac{(\frac23 (1.6 * 10^{-19}))^2}{3 * 10^{-15}}] + 2[(8.99 * 10^9) \frac{(\frac23 (1.6 * 10^{-19})\frac13(1.6 * 10^{-19}))}{3 * 10^{-15}}] [/tex]

    Then for the potential energy between both "up-quarks" I should be able to do:

    [tex]U = (8.99 * 10^9) \frac{(\frac23 (1.6 * 10^{-19}))^2}{3 * 10^{-15}}[/tex]

    Both of these are incorrect, I'm not understanding what I'm doing wrong?
     
    Last edited: Sep 24, 2015
  2. jcsd
  3. Sep 25, 2015 #2

    Orodruin

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    Why do you think it is wrong apart from the fact that you missed the sign of the down quark charge and that you are missing units?
     
  4. Sep 25, 2015 #3
    I was missing the sign! That was it. I always get confused about whether to include signs or take the absolute value/magnitude. In columb's law we don't use the signs do we? Every time I've tried to use negatives in coulumbs law it came out wrong, or maybe I'm just doing it wrong. That's why I assumed the negatives were not to be included here.
     
  5. Sep 25, 2015 #4

    Orodruin

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    This is something you really need to work on then. Usually you should include signs unless there is an explicit reason not to such as the thing being sought is the magnitude of the force. If you are taking absolute values on a hunch you are doing it wrong, you need to think through the reasons behind. (And you need to write out the units!)
     
  6. Sep 25, 2015 #5

    lightgrav

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    MeV !
     
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