Two electrons each moving with a velocity of `10^(6) ms^(-1)` are released towards eachother. What will be the closest distance of approach between them ?

Answer

`2.56xx10^(-10) m`

Explanation

`(1)/(2) mv^(2) + (1)/(2) mv^(2) = (e^(2))/(4pi in_(0) r)` , Calculate r.