Three point charges $q_{1}$ , $q_{2}$ and $q_{3}$ are taken such that when $q_{1}$ and $q_{2}$ are placed close together to form a single point charge, the force on $q_{3}$ at distance L from this combination is a repulsion of 2 unit in magnitude. When $q_{2}$ and $q_{3}$ are so combined the force on $q_{1}$ at distance L is an attractive force of magnitude 4 unit. Also $q_{3}$ and $q_{1}$ when combined exert an attractive force on $q_{2}$ of magnitude 18 unit at same distance L. The algebraic ratio of charges $q_{1}$ , $q_{2}$ , and $q_{3}$ is

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4 : -3 : 1

1 : 2 : 3

2 : -3 : 4

4 : -3 : 2

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{q_{3} (q_{1} + q_{2})}{L^{2}} = 2 \\ \frac{q_{1} (q_{2} + q_{3})}{L} = - 4 \\ \frac{q_{2} (q_{1} + q_{3})}{L^{2}} = - 18 \\ L e t \frac{q_{3}}{q_{1}} = K_{2} a n d \frac{q_{2}}{q_{1}} = K_{1} \\ \frac{K_{1} (K_{1} + 1)}{K_{1} (1 + K_{2})} = - \frac{1}{9} \\ \frac{K_{1} + K_{2}}{K_{2} (K_{1} + 1)} = - 2 \\ A f t e r s o l v i n g \\ K_{2} = \frac{2}{4} a n d K_{1} = \frac{- 3}{4} \\ Thus, q_{1} : q_{2} : q_{3} = 4 : - 3 : 2 \end{array}$