To solve this problem, we need to understand how the force between two charges changes when the charges are altered.$\\$Initially, the charges are $+Q$ and $-Q,$ and the force between them is given by Coulomb's law:$\\ F = \frac{k Q^2}{r^2} \\$where $k$ is Coulomb's constant and $r$ is the distance between the charges.$\\$When 25% of charge from A is transferred to B, the new charges become:$\\$• Charge on A: $Q_A = Q - 0.25Q = 0.75Q \\$• Charge on B: $Q_B = -Q + 0.25Q = -0.75Q \\$The new force $F'$ between the charges is:$\\ F' = \frac{k (0.75Q)(-0.75Q)}{r^2} \\$Simplify the expression:$\\ F' = \frac{k (0.75^2) Q^2}{r^2} \\ F' = \frac{k (0.5625) Q^2}{r^2} \\$Since the original force $F = \frac{k Q^2}{r^2},$ we can express $F'$ in terms of $F: \\ F' = 0.5625 \times F \\$Convert $0.5625$ to a fraction:$\\ 0.5625 = \frac{9}{16} \\$Thus, the new force $F' = \frac{9}{16} F. \\$Therefore, the correct option is $\frac{9F}{16},$ which corresponds to Option 2.