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) $\dfrac{2\sqrt{3}\,kq^2}{a^2}$
Each $+q$ attracts $-2q$ with $F = \dfrac{2kq^2}{a^2}$. The two forces make $60°$ with each other:
$$F_{net} = 2F\cos 30° = \sqrt{3}\cdot\frac{2kq^2}{a^2} = \frac{2\sqrt{3}\,kq^2}{a^2}$$
Solution by Sreeraj P, M.Sc Physics