A cyclotron accelerates two protons directly toward each other, each with a spee

Question

A cyclotron accelerates two protons directly toward each other, each with a speed of 1050 km/s, measured relative to the earth. Find the maximum electric force that these protons will exert on each other.


This is all the information I was given, and this is how I tried to solve. Can someone please work the problem? I'm out of time, and completely frazzled. All of my calculations have been wrong.

V = 1050 km/s Permittivity of free space (e0) = 8.85*10-12 Mproton= 1.67*10-27 Charge of electron (e) = 1.602*10-19 1. k = 1/4*pi*e0 = 1.3 * 1011 2. r = ke2/mv2 = 5.84 * 1021 3. Fmax = ke2/r2 = ? e0 V = 1050 km/s Permittivity of free space (e0) = 8.85*10-12 Mproton= 1.67*10-27 Charge of electron (e) = 1.602*10-19 1. k = 1/4*pi*e0 = 1.3 * 1011 2. r = ke2/mv2 = 5.84 * 1021 3. Fmax = ke2/r2 = ? e0

Explanation / Answer

The relative velocity v = v1 - v2 = 1.05 x 10^6 - ( -1.05 x10^6) = 2.10 x10^6 m/s K.E. of the proton K = (1/2)* m * v^2 = 0.5* 1.67 * 10^-27 * (2.10 *10^6)^2 = 3.68 x 10^-15 J Maximum force will be there when these protons are at minimum distance 'r' which happens when all the kinetic energy is converted into potential energy then (1/4pi e0) *q1 * q2 /r = 3.68 x 10^-15 r = (1/4pi e0)* q1 * q2 / 3.68 x 10^-15 (e stands for epsilon) Hence maximum force F = (1/4pi e0) * q1 * q2 / r2 = (1/4pi e0) * q1 * q2 / {(1/4pi e0) *q1 * q2 / 3.68 x 10^-15 }^2 = (3.68 x 10^-15)^2 / (1/4pi e0)* 1.6 * 10^-19 * 1.6* 10^-19 = 1.35* 10-29 / 9 * 10^9 * 2.56 *10-38 = 0.058 N

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