• # question_answer A charge q is placed at the centre of the joining the equal charges Q. The system of three charges will be in equilibrium if q is equal to A) $\frac{-Q}{2}$                                 B) $\frac{-Q}{4}$C) $-4Q$                        D) $\frac{+Q}{2}$

Let the distance between the two charges Q and Q be r. Force on Q due to charge q at a distance (r/2) $=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{qQ}{{{(r/2)}^{2}}}$ Force on Q due to other charge Q at a distance r $=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{Q+Q}{{{(r)}^{2}}}$ $\therefore$$\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{qQ}{{{(r/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{Q\times Q}{{{r}^{2}}}=0$ $4q=-Q$or $q=-Q/4$ Hence, the correction option is [b].