Answer
$v=22\frac{km}{s}$
Work Step by Step
In equilibrium state,
$K.E=P.E$
That is,
$\frac{1}{2}mv^2=K\frac{qQ}{r}$
$v=\sqrt\frac{2KqQ}{rm}$
We plug in the known values in the formula to obtain:
$v=\sqrt\frac{2\times 9\times 10^9\times1.6\times 10^{-19}\times 1.6\times 10^{-15}}{0.01\times 9.109\times 10^{-31}}$
$v=22492$
$v=22\frac{km}{s}$
