Two charges 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

Suppose distance AC = x and AB = d Repulsive force on 2q by q,
`E_(q) = (k(q)(2q))/x^(2)`……(1)
Attractive force on 2q by `-3q`
`E_(-3q) = (k(2q)(3q))/(x+d)^(2)`………..(2)
Resultant force on 2q.
`F = Fq + F_(-3q)`
(`therefore` No force on 2q)
`therefore Fq =-F_(-3q)`
`k(2q^(2))/x^(2) = +k(6q^(2))/(x+d)^(2)`
`therefore 1/x^(2) =3/(x+d)^(2)`
By taking square root on both sides,
`1/x = sqrt(3)/(x|d) = 1.732/(x+d)`
`therefore x+d = 1.732 x`
`therefore x +d = 1.732 x`
`therefore x = 1.366 d`
`=d/2(1+sqrt(3))`