Question

Two protons, starting several meters apart, are aimed directly at each other with speeds of 2.00 * 105 m>s, measured relative to the earth. Find the maximum electric force that these protons will exert on each other.

Answer 1

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.

           $U_{a} + K_{a} = U_{b} + K_{b}$

or,           $K_{a} = U_{b}$  

where,   $K_{a}$ = kinetic energy of two protons

So,      $2(\frac{1}{2}mV^{2}_{a}) = \frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r_{b}}$

       $r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

Putting the given values into the above formula we will calculate the value of $r_{b}$ as follows.

      $r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

                = $(9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}^{2} C}{1.67 \times 10^{-27} kg \times 2 \times 10^{5}}$

                = $3.45 \times 10^{-12} m$

Now, we will calculate the maximum electric field as follows.

      F = $\frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r^{2}_{b}}$

         = $9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}C)^{2}}{3.45 \times 10^{-12}}$

         = $0.194 \times 10^{-4} N$

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