Question

Two identically charged one-peso coins are 1.5 m apart on a table. What is the charge of the coins if each of them experiences a repulsive force of 2.0 N?

Answer 1

The Coulomb's law says

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

The absolute value of charge

$$q=\sqrt{Fr^2/k}\\ q=\sqrt{\frac{2.0*1.5^2}{9*10^9}}=2.2*10^{-5}\:\rm C$$