Electric Charge & Coulomb's Law question from MHT CET 2025 (25 Apr, Shift 2)

Answer: (b)

1. Initially $F = \dfrac{kq^2}{r^2}$. After touching, the third ball and the touched ball each have $q/2$.
2. At the midpoint (distance $r/2$ from each): force from the untouched ball $= \dfrac{kq(q/2)}{(r/2)^2} = 2F$.
3. Force from the touched ball $= \dfrac{k(q/2)(q/2)}{(r/2)^2} = F$, in the opposite direction.
4. Net force $= 2F - F = F$. So the answer is (b) $F$.