Rajat
Last Activity: 5 Years ago
Let the charges q be placed on A and B and q' be placed on C. O is the origin.
Let angle ACO= x
So, AC =BC = a cosecx
Force on q' by each charge = kqq'/a^2cosec^2x
Vector sum of the total force acting on charge q' =2(kqq'/a^2cosec^2x)cosx
= (2kqq'/a^2)sin^2xcosx
Enough to maximise sin^2xcosx
Differentiating it we get sinx(3cos^2x-1) = 0
So, x = 0 ( not possible since q' have to be at infinity)
And cosx= 1/√3
So, y/√(y^2+a^2) = 1/√3
Or, (y^2+a^2)/y^2 = 3
Or, (a/y)^2= 2
So, y = a/√2
