Answer
$v_f=2.5 \frac{Km}{s}$
Work Step by Step
We know that:
$dK=-dU$
$\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=-(\frac{kq^2}{r_f}-\frac{kq^2}{r_i})$
We plug in the known values to obtain:
$\frac{1}{2}(20\times 10^{-6})\times (v_f^2)-\frac{1}{2}(20\times 10^{-6})\times (0)=-(\frac{9\times 10^9\times (3.1\times 10^{-6})^2}{2.5\times 10^{-3}}- \frac{9\times 10^9\times (3.1\times 10^{-6})^2}{0.90\times 10^{-3}})$
Thus;
$v_f=2500\frac{m}{s}=2.5 \frac{Km}{s}$
