In the quark model of fundamental particles, a proton is composed of three quarks: two “up” quarks, each having charge $\mathbf{+}\mathbf{2}\mathit{e}\mathbf{/}\mathbf{3}$, and one “down” quark, having charge $\mathbf{-}\mathit{e}\mathbf{/}\mathbf{3}$. Suppose that the three quarks are equidistant from one another. Take that separation distance to be $\mathbf{1}\mathbf{.32}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{\text{m}}$and calculate the electric potential energy of the system of

(a) only the two up quarks and

(b) all three quarks.

The charge,$Q_1=Q_2=\frac{+2e}{3}$

The charge,$Q_3=\frac{-e}{3}$

The electric charge,$e=1.6\times {10}^{-19}\text{C}$

The distance, $R=1.32\times {10}^{-15}\text{m}$

After reading the question, the potential energy of two charged system is,

$\mathit{U}\mathbf{=}\frac{\mathbf{k}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{R}}$

If two like charges (two protons or two electrons) are brought towards each other, the potential energy of the system increases. If two unlike charges i.e. a proton and an electron are brought towards each other, the electric potential energy of the system decreases.

Formula:

Write the equation for the potential energy as below.

$U=\frac{k{Q}_1{Q}_2}{R}$

Where,K is the coulomb constant having a value ${\text{9.0×10}}^{\text{9}}\text{N}\cdot \text{m}/{\text{C}}^{\text{2}}$, $Q_1$ and $Q_2$ are the charges in the system, and R is the distance between the charges.

The 3 charges are given to be equidistant from each other. The electric potential of only two up quarks will be due to the interaction of two up quarks.

$U=\frac{k{Q}_1{Q}_2}{R}$

Substitute$\frac{+2e}{3}$ for $Q_1$and, ${\text{9.0×10}}^{\text{9}}\text{N}\cdot \text{m}/{\text{C}}^{\text{2}}$ for K, $Q_2$and ${\text{1.32×10}}^{\text{-15}}\text{m}$for R in the above equation.

$\begin{array}{c}U=\frac{k\left(\frac{+2e}{3}\right)\left(\frac{+2e}{3}\right)}{R}\\ =\frac{k(+4e)}{9R}\end{array}$

$\begin{array}{c}U=\frac{\text{4}(9.0\times {10}^9\text{N}\cdot \text{m}/{\text{C}}^{\text{2}}){(1.6\times {10}^{-19}\text{C})}^{\text{2}}}{\text{9}(1.32\times {10}^{-15}\text{m})}\\ =7.76\times {10}^{-14}\text{J}\end{array}$

Hence, the potential energy of two system of two quarks is $7.76\times {10}^{-14}\text{J}$.

The potential energy of the system with all three quarks is,

$U=U_{12}+U_{23}+U_{13}$ ….. (1)

Here,is the potential energy due to interaction of 2 up quarks$\frac{+2e}{3}$and$\frac{+2e}{3}$, $U_{23}$is the potential energy due to interaction of second up quark$\frac{+2e}{3}$and the down quark$\frac{-e}{3}$and$U_{13}$is the potential energy due to interaction of first up quark$\frac{+2e}{3}$and the down quark$\frac{-e}{3}$.

$\begin{array}{c}U=\frac{k\left(\frac{+2e}{3}\right)\left(\frac{+2e}{3}\right)}{1.32\times {10}^{-15}\text{m}}+\frac{k\left(\frac{+2e}{3}\right)\left(\frac{-e}{3}\right)}{1.32\times {10}^{-15}\text{m}}+\frac{k\left(\frac{+2e}{3}\right)\left(\frac{-e}{3}\right)}{1.32\times {10}^{-15}\text{m}}\\ =\frac{4e^{2}k}{9\times 1.32\times {10}^{-15}\text{m}}-\frac{2k{e}^2}{9\times 1.32\times {10}^{-15}\text{m}}-\frac{2k{e}^2}{9\times 1.32\times {10}^{-15}\text{m}}\\ =\frac{k\times e^2}{9(1.32\times {10}^{-15}\text{m})}(4-2-2)\\ =0\text{J}\end{array}$

Hence, the net potential energy of all three quarks is $0\text{J}$.