Arun
Last Activity: 6 Years ago
Let Q1 be a test positive charge, since charges of Q2 and Q3 are opposite in sign, there exists an attractive force between them. One point to be kept in mind here is that the net force on Q3 must be 0 as given in the problem. In order to keep force on Q3 be 0, the force between Q1 and Q3 must be force of repulsion as we already have attractive force between Q2 and Q3Now to satisfy the equilibrium position, the two forces must balance each other.KQ1Q3/4R2 = -KQ2Q3/R2….. (distance between Q1 and Q3 is 2R).Q1 = -4Q2Hence charge of Q1 is opposite to that of Q2 and same as that of Q3