Respuesta :
Explanation:
Since, the two protons are at the initial state of point a and point b. Hence, total mechanical energy at point a and point b is as follows.
[tex]U_{a} + K_{a} = U_{b} + K_{b}[/tex]
or, [tex]K_{a} = U_{b}[/tex]
where, [tex]K_{a}[/tex] = kinetic energy of two protons
So, [tex]2(\frac{1}{2}mV^{2}_{a}) = \frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r_{b}}[/tex]
[tex]r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}[/tex]
Putting the given values into the above formula we will calculate the value of [tex]r_{b}[/tex] as follows.
[tex]r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}[/tex]
= [tex](9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}^{2} C}{1.67 \times 10^{-27} kg \times 2 \times 10^{5}}[/tex]
= [tex]3.45 \times 10^{-12} m[/tex]
Now, we will calculate the maximum electric field as follows.
F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r^{2}_{b}}[/tex]
= [tex]9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}C)^{2}}{3.45 \times 10^{-12}}[/tex]
= [tex]0.194 \times 10^{-4} N[/tex]
Therefore, we can conclude that the maximum electric force that these protons will exert on each other is [tex]0.194 \times 10^{-4} N[/tex].
