## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$q_1 = -20~nC$
Let $F_L$ be the magnitude of the force on the charge $q_2$ (directed to the left) from $q_1$. Let $F_R$ be the magnitude of the force on the charge $q_2$ (directed to the right) from the $5.0~nC$ charge. Since $q_2$ is in static equilibrium, $F_L = F_R$. We can find the magnitude of the charge $q_1$: $F_L = F_R$ $\frac{k~q_1~q_2}{(2r)^2} = \frac{k~(5.0~nC)~q_2}{r^2}$ $\frac{q_1}{4} = 5.0~nC$ $q_1 = 20~nC$ Since the charge on $q_1$ is negative, $q_1 = -20~nC$.