91Ó°ÊÓ

The electric force between two protons is as follows:

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}_{\mathbf{b}}^{\mathbf{2}}}$

The electric force acting of protons is derived as follows:

role="math" localid="1664256985484" $\mathrm{F}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\left|{\mathrm{q}}_1{\mathrm{q}}_2\right|}{{\mathrm{r}}_{\mathrm{b}}^2}$

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}_{\mathbf{b}}^{\mathbf{2}}}$

Here,role="math" localid="1664256998618" $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}×9×{10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}$

Consider the charge of the proton is $1.602×{10}^{-19} C$.

Consider the total kinetic energy for those two particles are equal.

$$\begin{gathered} 2\left(\frac{1}{2}m{v}^2\right)=\frac{1}{4\mathrm{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\left|e^2\right|}{r_b} \\ {r}_b=\frac{1}{4\mathrm{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{e^2}{m{v}^2} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} {\mathrm{r}}_{\mathrm{b}}=\left(9×{10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}\right)\frac{{\left(1.602×{10}^{-19}\mathrm{C}\right)}^2}{\left(1.67×{10}^{-27}\mathrm{kg}\right){\left(2×{10}^5{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)}^2} \\ =3.45×{10}^{-12}\mathrm{m}, \end{gathered}$$

Solve for the maximum force as:

$$\begin{gathered} \mathrm{F}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\left|{\mathrm{e}}^2\right|}{{\mathrm{r}}_{\mathrm{b}}^2} \\ =\left(9×{10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}\right)\frac{{\left(1.607×{10}^{-19}\mathrm{C}\right)}^2}{{\left(3.45×{10}^{-12}\mathrm{m}\right)}^2} \\ =0.194×{10}^{-4}\mathrm{N} \end{gathered}$$

Therefore, the maximum electric force is$0.194×{10}^{-4} \mathrm{N}$ that these protons will exert on each other.