Question

A neutron consists of one “up” quark of charge $\mathbf{+}\mathbf{2}\mathbf{e}\mathbf{/}\mathbf{3}$and two “down” quarks each having charge$\mathbf{-}\mathit{e}\mathbf{/}\mathbf{3}$. If we assume that the down quarks are$\mathbf{2.6}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathit{m}$apart inside the neutron, what is the magnitude of the electrostatic force between them?

Short answer

The magnitude of electrostatic force between them is$3.8N$

Step-by-step solution

The given data

1. A neutron consists of one “up” quark of charge$+2\mathrm{e}/3$ and two “down” quarks of charge$-e/3$
2. Distance between the two down quarks,$d=2.6\times {10}^{-15}m$

Understanding the concept of Coulomb’s law

Using the concept of Coulomb's law, we can find the magnitude of the required force between the particles. Now, equating this net force to zero, we can get the value of the charge.

Formula:

The magnitude of the electrostatic force between any two particles, $\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$(i)

Calculation of the magnitude of the force

The electrostatic force between the two down quarks using equation (i) and the given data is given as:

$\begin{array}{c}\mathrm{F}=\mathrm{k}\frac{{(\mathrm{e}/3)}^2}{{\mathrm{r}}^2}\\ =\frac{(9.00\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2){(1.60\times {10}^{-19}\mathrm{C})}^2}{9{(2.6\times {10}^{-15}\mathrm{m})}^2}\\ =3.8\mathrm{N}\end{array}$

Hence, the value of the force is $3.8\mathrm{N}$