Two electrons are moving towards each other,each with a velocity of 10^{6}m/s.What will be losest distance of approach between them?

$K.E=P.E\phantom{\rule{0ex}{0ex}}2\times \frac{1}{2}m{v}^{2}=\frac{k\times e\times e}{r}herem=massofelectron,k=9\times {10}^{9}N-{m}^{2}/{C}^{2}\phantom{\rule{0ex}{0ex}}=r=\frac{k\times {e}^{2}}{m{v}^{2}}=\frac{9\times {10}^{9}\times (1.6\times {10}^{-19}{)}^{2}}{9.1\times {10}^{-31}\times ({10}^{6}{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9\times 2.56\times {10}^{9}\times {10}^{-38}}{9.1\times {10}^{-31}\times {10}^{12}}=\frac{2.53\times {10}^{-29}}{{10}^{-19}}\phantom{\rule{0ex}{0ex}}=2.53\times {10}^{-29+19}\phantom{\rule{0ex}{0ex}}=2.53\times {10}^{-10}meter.$

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