Question

an electric force 3.24x10^-4 C exerts an attractive force of 12 N on a second charge that is 0.040 m away. what is the magnitude of the second charge?

Answer 1

According to Coulomb's law:

$$F=k\frac{q_1q_2}{r^2},\\\space\\ q_2=\frac{r^2F}{kq_1}=\frac{0.04^2·12}{(9·10^9)(3.24·10^{-4})}=6.58·10^{-9}\text{ C}.$$