Two identical particles of mass m and charge | vedclass

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(B) At the distance of closest approach,the relative velocity of the particles becomes zero. By the law of conservation of energy,the total initial ki

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$\frac{q^2}{4 \pi \varepsilon_0 m v^2}$

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(B) At the distance of closest approach,the relative velocity of the particles becomes zero. By the law of conservation of energy,the total initial kinetic energy is equal to the total potential energy at the distance of closest approach $r$.The total initial kinetic energy is $KE_i = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2$.The potential energy at distance $r$ is $PE_f = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}$.Equating the two: $m v^2 = \frac{q^2}{4 \pi \varepsilon_0 r}$.Solving for $r$,we get $r = \frac{q^2}{4 \pi \varepsilon_0 m v^2}$.

Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

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