Q 12-01-011NEETJEE MainMedium
Charges $+q$, $+q$ and $-2q$ are placed at the corners of an equilateral triangle of side $a$. The magnitude of the net force on the $-2q$ charge is
Answer: (C) $\dfrac{2\sqrt{3}\,kq^2}{a^2}$
Concept. Add the two attractive forces on $-2q$ as vectors; they are equal and $60°$ apart.
Step 1. Each force. $F = \dfrac{k(q)(2q)}{a^2} = \dfrac{2kq^2}{a^2}$.
Step 2. Resultant.
$$F_{\text{net}} = 2F\cos30° = \sqrt3 \cdot \frac{2kq^2}{a^2} = \frac{2\sqrt3\,kq^2}{a^2}$$
Answer. $\dfrac{2\sqrt3\,kq^2}{a^2}$ (Option C).
Solution by Sreeraj P, M.Sc Physics