Answer :
Answer:
[tex]1.36\cdot 10^{-4} N[/tex]
Explanation:
To keep the problem simpler, we can imagine that one proton is at rest and the second proton is approaching the first one with a speed which is twice the speed given, therefore:
[tex]v=2\cdot 2.30\cdot 10^5 m/s=4.30\cdot 10^5 m/s[/tex]
According to the law of conservation of energy, the second proton will stop when all its kinetic energy has been converted into electric potential energy, therefore when
[tex]q V = \frac{1}{2}mv^2[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton's charge
V is the electric potential at the final location of the proton
[tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass
The electric potential generated by the first proton is
[tex]V=\frac{kq}{r}[/tex]
where
[tex]k=9\cdot 10^9 Nm^2 C^{-2}[/tex] is the Coulomb's constant
r is the minimum distance of approach of the second proton
Substituting all the quantities and solving for r, we find:
[tex]k\frac{q^2}{r}=\frac{1}{2}mv^2\\r=\frac{2kq^2}{mv^2}=\frac{2(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(1.67\cdot 10^{-27})(4.6\cdot 10^5)^2}=1.30\cdot 10^{-12} m[/tex]
And so, the maximum electrical force between the two protons will be:
[tex]F=k\frac{q^2}{r^2}=(9\cdot 10^9) \frac{(1.6\cdot 10^{-19})^2}{(1.30\cdot 10^{-12})^2}=1.36\cdot 10^{-4} N[/tex]
The maximum electrical force on the protons will be [tex]1.36\times10^{-4}N[/tex]
Given that two protons moving towards each other with the same velocity [tex]u = 2.30\times10^5m/s[/tex]
Relative speed:
The speed of one proton relative to the other is given by:
[tex]v=2u=4.6\times10^5m/s[/tex]
The protons will repel each other so they will stop at a distance from each other due to repulsion, the distance is called minimum distance of approach.
Conservation of energy:
According to the law of conservation of energy, the electrostatic potential energy between the protons at the minimum distance of approach r must be equal to the kinetic energy of the protons
[tex]k\frac{q^2}{r}=\frac{1}{2}mv^2\\ \\ r=\frac{2kq^2}{mv^2}=\frac{2*9*10^{9}*(1.6*10^{-19})^2}{9.1*10^{-31}*4.6*10^5} \\\\r=1.3\times 10^{-12}m[/tex]
Now the electrostatic force is
[tex]F=k\frac{q^2}{r^2}=9*10^9*\frac{(1.6*10^{-19})^2}{(1.3*10^{-12})^2}\\ \\ F=1.36\times10^{-4}N[/tex]is the force at the minimum distance of approach of the protons.
Learn more about electrostatic force:
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