Question

given two charges q1=3x10^-6C and q2=-4x10^-6 C find the potential emergy of the system when they are 2x10^-6 cm apart

Answer 1

$$U=-k\frac{q_1q_2}{r}$$

$$U=-9\cdot10^9\frac{(3\cdot10^{-6})(-4\cdot10^{-6})}{2\cdot10^{-6}}=54000\ J$$