Question

1. two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1000 km/s, measured relative to the earth. find the maximum electrical force that these protons will exert on each other.

Explanation:

Step1: Apply conservation of energy

The initial kinetic energy of the two protons is converted into electrical potential energy at the point of closest approach. The initial kinetic energy \(K_{i}\) of each proton is \(K=\frac{1}{2}mv^{2}\), and for two protons \(K_{i} = 2\times\frac{1}{2}mv^{2}=mv^{2}\). The electrical potential energy is \(U = k\frac{q_{1}q_{2}}{r}\), where \(q_{1}=q_{2}=e = 1.6\times10^{-19}\space C\), \(k = 9\times10^{9}\space N\cdot m^{2}/C^{2}\). At the point of closest approach, \(K_{i}=U\).

Step2: Calculate the velocity conversion

The speed \(v = 1000\space km/s=10^{6}\space m/s\), and the mass of a proton \(m = 1.67\times 10^{-27}\space kg\). From \(mv^{2}=k\frac{e^{2}}{r}\), we can solve for \(r\): \(r = k\frac{e^{2}}{mv^{2}}\).

Step3: Calculate the electrical force

The electrical force \(F=k\frac{e^{2}}{r^{2}}\). Substituting \(r = k\frac{e^{2}}{mv^{2}}\) into the force formula, we get \(F=\frac{m^{2}v^{4}}{k e^{2}}\).

Substitute \(m = 1.67\times 10^{-27}\space kg\), \(v = 10^{6}\space m/s\), \(k = 9\times10^{9}\space N\cdot m^{2}/C^{2}\), \(e = 1.6\times10^{-19}\space C\) into the formula:

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Answer:

The maximum electrical force is \(0.121\space N\)