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## Homework Statement

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?

## Homework 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]

## The Attempt at a Solution

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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?

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