RRB Technician Grade 1 Model Paper 3

Concept:For equilibrium of the system, the net force on every charge must be zero.The charge $q$ must be negative so that it attracts the two outer charges $Q$ and balances their mutual repulsion.Explanation:Let the two equal charges $Q$ be placed at points A and B, with distance $l$ between them.The charge $q$ is at the midpoint C, so $AC = BC = \frac{l}{2}$.Consider the equilibrium of the charge $Q$ at A.The repulsive force on A due to the other charge $Q$ at B is:$$F_{AB} = K\frac{Q^2}{l^2}$$The force on A due to charge $q$ at C is:$$F_{AC} = K\frac{Q q}{\left(\frac{l}{2}\right)^2} = K\frac{4Qq}{l^2}$$For equilibrium, these forces must be equal in magnitude and opposite in direction.Since $F_{AB}$ is repulsive, $F_{AC}$ must be attractive, so $q$ must be negative.Equating magnitudes:$$K\frac{4Q|q|}{l^2} = K\frac{Q^2}{l^2}$$$$4|q| = Q$$$$|q| = \frac{Q}{4}$$Hence, $q = -\frac{Q}{4}$.The central charge $q$ also experiences equal and opposite forces from the two charges $Q$, so it is already in equilibrium.Answer:Option B: $-\frac{Q}{4}$