06 \ \mu m = 1.06 \times 10^{-6}$ m. The final distance between the charges is $r_f = 0.340 \ \mu m = 0.340 \times 10^{-6}$ m.
The initial potential energy is $U_i = \frac{k q_1 q_2}{r_i}$, where $k = 8.99 \times 10^9 \ N m^2/C^2$.
$U_i = \frac{(8.99 \times 10^9 \
Show more…