Question

Two equal charges are placed 100m a part of the force of attraction between them is 0.2N,calculate the value of the two equal charges

Answer 1

$$F=k\frac{q_1q_2}{r^2}=k\frac{q^2}{r^2}$$

$$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{0.2*100^2}{9*10^9}}=4.7*10^{-4}\:\rm C$$