Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% of charge of A is transferred to B, then the new force between A and B is?

Correct option is B.
9F/16
F=KQ²/r²
If 25% of charge of A is transferred to B, then charge on A becomes Q - 0.25Q = 0.75Q and charge on B becomes -Q + 0.25Q = -0.75Q.
The new force F' = K(0.75Q)(-0.75Q)/r² = 0.5625 KQ²/r² = 0.5625F
Since F = KQ²/r², F' = 0.5625F = (9/16)F