Two charge q and `–3q` are placed fixed on x–axis separated by distance d. Where should a third charge` 2q` be placed such that it will not experience any force ?

Explanation

Here, let us keep the charge 2q at a distance r from A.

Thus, charge 2q will not experience any force.
when, force of repulsion on it due to q is balanced by force of attraction on it due to -3 q, at B, where AB=d.
thus, force of attraction by -3q=force of repulsion by q
`implies(2qxxq)/(4piepsi_(0)x^(2))=(2qxx3q)/(4piepsi_(0)(x+d)^(2))`
`implies(x+d)^(2)=3x^(2)` ltBrgt `impliesx^(2)+d^(2)+3xd=3x^(2)`
`implies=2x^(2)-d^(2)` ltBrgt `therefore2x^(2)-2dx-d^(2)=0` ltBrgt `x=(d)/(2)+-(sqrt(3)d)/(2)` (Negative sigh be between q and -3q and hence is unadaptable.)
`x=-(d)/(2)+(sqrt(3)d)/(2)`
`=(d)/(2)(1+sqrt(3))` to the left of q.