Question

In the quark model of fundamental particles, 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 1.20×10^-15 m and calculate the potential energy of the subsystem of two "up" quarks. (MeV)

Answer 1

The potential energy of a system formed by two charge is

$U=K\frac{q_{1}q_{2}}{r_{12}}$

where $r_{12}$   is the distance between the charges. Therefore, if

$q_{1}=q_{2}=\tfrac{2}{3}e=\mathrm{\tfrac{2}{3}\left( 1.6\times10^{-19}\,C \right)=1.07\times10^{-19}\,C}$

the potential energy is

$U=K\frac{q_{1}q_{1}}{r_{12}}$

$U=K\frac{q_{1}^{2}}{r_{12}}=K\frac{q_{2}^{2}}{r_{12}}$

$U=\mathrm{\left( 9\times10^{9}\,\tfrac{N.m^{2}}{C^{2}} \right )\frac{\left( 1.07\times10^{-19}\,C \right)^{2}}{\left( 1.20\times10^{-15}\,m \right)}}$

$U=\mathrm{8.59\times10^{-14}\,J}$

but,

$\mathrm{1\,eV=1.6\times10^{-19}\,J} \begin{matrix} && --> && \end{matrix} \mathrm{1\,J=\frac{1}{1.6\times10^{-19}}\,eV}$

finally

$U=\mathrm{8.59\times10^{-14}\,J}$

$U=\mathrm{8.59\times10^{-14}\left( \frac{1}{1.6\times10^{-19}}\,eV \right)}$

$U=\mathrm{0.537\,MeV}$