Two electrons are moving towards each other, each with a velocity of $10^6 \,m / s$. What will be closest distance of approach between them is ......... $m$
Medium
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(b)
$2 . \frac{1}{2}\left(9.1 \times 10^{-31}\right)\left(10^6\right)^2=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{r}$
$9.1 \times 10^{-19}=\frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{r}$
$r=2.56 \times 10^{-10} \,m$
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