Electric Charge & Coulomb's Law question from JEE Main 2022 (27 Jun, Shift 2)

Answer: (d)

1. Force between any two balls: $F = \dfrac{kq^2}{(1)^2}$.
2. On one ball, the other two balls exert two forces of magnitude $F$ with $60^\circ$ between them.
3. Net electric force $= \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{3}F$
4. Ratio $= \sqrt{3}F : F = \sqrt{3} : 1$
5. So the answer is (d) $\sqrt{3} : 1$.