Question

Suppose we have one moving electron ,approaching towards a fixed electron. If they 1mm apart when the moving electron has the velocity of 1000cm/s. How far will be the moving electron from the fixed electron when it comes to the rest.
mass of electron=9.1E-31 Kg
charge of electron =1.6E-19 C

Answer 1

initial velocity (u)= 1000 cm/s = 10 m/s
final velocity(v) = 0
initial distance ($r_1$)= 1 mm = 0.001 m
final distance = $r_2$
mass of each electron (m) = 9.1E-31 Kg
charge (q) on each electron = -1.6E-19 C

According to conservation of energy, the change in KE of the moving electron is equal to the increase in PE of the system. So 

$\Delta KE=\Delta PE\\ \\ \frac{1}{2} m(v^{2}-u^2)= \frac{q}{4 \pi \epsilon_0} ( \frac{1}{r_1} - \frac{1}{r_2} )\\ \\ \frac{1}{2} *9.1*10^{-31}(0-10^2)= (9*10^{9})(-1.6*10^{-19}) ( \frac{1}{0.001} - \frac{1}{r_2} )\\ \\\frac{1}{2} *9.1*10^{-31}*100= (-9*1.6*10^{-10}) ( 1000 - \frac{1}{r_2} )\\ \\ \frac{1}{2} *9.1*10^{-29}= -1.44*10^{-10}* ( 1000-\frac{1}{r_2} )\\ \\1000-\frac{1}{r_2}=- \frac{0.5*9.1*10^-{29}}{1.44*10^{-10}} =-3.16*10^{-19}\\ \\ \frac{1}{r_2}=1000+ 3.16*10^{-19}$

$r_2= \frac{1}{1000+ 3.16*10^{-19}}$

$r_2$ will be slightly less than 1 mm.