## Fundamentals of Physics Extended (10th Edition)

$v_f=2.5 \frac{Km}{s}$
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; $v_f=2500\frac{m}{s}=2.5 \frac{Km}{s}$