Import Question JSON

$\text{Hint: } U = \frac{kq_1q_2}{r}$ $\text{Step: Find the total potential energy of the system.}$ $\text{The potential energy } U \text{ between two charges } q_1 \text{ and } q_2 \text{ separated by a distance given by; } U = \frac{kq_1q_2}{r}$ $\text{The potential energy for all pairs of charges in this configuration.}$ $\text{The potential energy between } -Q \text{ and } -2Q \text{ is given by;}$ $U_1 = \frac{k(-Q)(-2Q)}{l} = \frac{2kQ^2}{l}$ $\text{The potential energy between } -Q \text{ and } q \text{ is given by; (Since, } q \text{ is at the midpoint, the distance between } -Q \text{ and } q \text{ is } \frac{l}{2})$ $U_2 = \frac{k(-Q)(q)}{l/2} = -\frac{2kQq}{l}$ $\text{The potential energy between } -2Q \text{ and } q \text{ is given by; (the distance between } -2Q \text{ and } q \text{ is } \frac{l}{2})$ $U_3 = \frac{-2kQq}{l/2} = -\frac{4kQq}{l}$ $\text{Now, sum all the contributions:}$ $U_{\text{total}} = U_1 + U_2 + U_3$ $U_{\text{total}} = \frac{2kQ^2}{l} - \frac{2kQq}{l} - \frac{4kQq}{l}$ $U_{\text{total}} = \frac{2kQ^2}{l} - \frac{6kQq}{l}$ $U_{\text{total}} = \frac{kQ}{l}(2Q - 6q)$ $\text{For the total potential energy to be positive:}$ $2Q - 6q > 0$ $2Q > 6q$ $\Rightarrow q < \frac{Q}{3}$ $\text{The system will have a positive potential energy configuration if, } q < \frac{Q}{3}.$ $\text{Hence, option (2) is the correct answer.}$