$(a)$ In a quark model of elementary particles,a neutron is made of one up quark [ charge $\frac{2}{3}e$ ] and two down quarks [ charges $-\frac{1}{3}e$ ]. Assume that they have a triangle configuration with side length of the order of ${10^{ - 15}} \ m$. Calculate the electrostatic potential energy of the neutron and compare it with its mass $939 \ MeV$. $(b)$ Repeat the above exercise for a proton which is made of two up and one down quark.

(N/A) The electrostatic potential energy of a system of three charges is given by $U = k \left( \frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_3 q_1}{r} \right)$.
For a neutron,the charges are $q_1 = \frac{2}{3}e$,$q_2 = -\frac{1}{3}e$,and $q_3 = -\frac{1}{3}e$.
$U = \frac{k}{r} \left[ (\frac{2}{3}e)(-\frac{1}{3}e) + (-\frac{1}{3}e)(-\frac{1}{3}e) + (-\frac{1}{3}e)(\frac{2}{3}e) \right] = \frac{k}{r} \left[ -\frac{2}{9}e^2 + \frac{1}{9}e^2 - \frac{2}{9}e^2 \right] = \frac{k}{r} \left( -\frac{3}{9}e^2 \right) = -\frac{k e^2}{3r}$.
Substituting $k = 9 \times 10^9 \ N \ m^2/C^2$,$e = 1.6 \times 10^{-19} \ C$,and $r = 10^{-15} \ m$:
$U = -\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{3 \times 10^{-15}} = -7.68 \times 10^{-14} \ J$.
Converting to $eV$: $U = \frac{-7.68 \times 10^{-14}}{1.6 \times 10^{-19}} \ eV = -4.8 \times 10^5 \ eV = -0.48 \ MeV$.
$(b)$ For a proton,the charges are $q_1 = \frac{2}{3}e$,$q_2 = \frac{2}{3}e$,and $q_3 = -\frac{1}{3}e$.
$U = \frac{k}{r} \left[ (\frac{2}{3}e)(\frac{2}{3}e) + (\frac{2}{3}e)(-\frac{1}{3}e) + (-\frac{1}{3}e)(\frac{2}{3}e) \right] = \frac{k}{r} \left[ \frac{4}{9}e^2 - \frac{2}{9}e^2 - \frac{2}{9}e^2 \right] = 0 \ J$.