Answer
$v=3.9\times 10^3\frac{m}{s}$
Work Step by Step
In the given scenario, all potential energy is converted into kinetic energy.
According to law of conservation of energy:
$PE_{initial }=KE_{final}$
$\implies \frac{KQ^2}{r}=2(\frac{1}{2}mv^2)$
This simplifies to
$v=\sqrt{\frac{KQ^2}{mr}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{9\times 10^9\times (9.5\times 10^{-6})^2}{1.0\times 10^{-6}\times 5.3\times 10^{-2}}}=3.9\times 10^3\frac{m}{s}$
