91影视

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

The charge,$Q_3=\frac{鈭�e}{3}$

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

The distance, $R=1.32脳{10}^{鈭�15}\text{m}$

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 ${\text{9.0脳10}}^{\text{9}}\text{N}鈰�\text{m}/{\text{C}}^{\text{2}}$, $Q_1$ 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, ${\text{9.0脳10}}^{\text{9}}\text{N}鈰�\text{m}/{\text{C}}^{\text{2}}$ for K, $Q_2$and ${\text{1.32脳10}}^{\text{-15}}\text{m}$for R in the above equation.

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

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

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

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

$U=U_{12}+U_{23}+U_{13}$ 鈥�.. (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$\frac{鈭�e}{3}$and$U_{13}$is the potential energy due to interaction of first up quark$\frac{+2e}{3}$and the down quark$\frac{鈭�e}{3}$.

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

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