Question

Two protons, starting several meters apart, are aimed directly at each other with speeds of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\frac{\mathbf{m}}{\mathbf{s}}$, measured relative to the earth. Find the maximum electric force that these protons will exert on each other.

Short answer

The maximum electric force is$0.194\times {10}^{-4}\hspace{0.33em}\mathrm{N}$ that these protons will exert on each other.

Step-by-step solution

Electric force

The electric force between two protons is as follows:

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}_{\mathbf{b}}^{\mathbf{2}}}$

The electric force acting of protons is derived as follows:

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

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}_{\mathbf{b}}^{\mathbf{2}}}$

Here,role="math" localid="1664256998618" $\frac{1}{4\mathrm{\pi }{\epsilon }_0}\times 9\times {10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}$

Determine the value of electric force

Consider the charge of the proton is $1.602\times {10}^{-19}\hspace{0.33em}C$.

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

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

Substitute the values and solve as:

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

Solve for the maximum force as:

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

Therefore, the maximum electric force is$0.194\times {10}^{-4}\hspace{0.33em}\mathrm{N}$ that these protons will exert on each other.