Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieveequilibrium for the entire system and where should it be placed form charge q? Why we do not equate equilibrium at Q rather than at q or 4q for magnitude of Q

4qandqare given to be free positive charges placed at a distancelapart. So, chargeQneeded to achieve equilibrium for the entire system, should be of negative sign. LetQis at a distancexfromqcharge, and therefore(l−x)from4qcharge.
From Coulomb's Law, force between two charges isr2kq1​q2​​
So, for equilibrium, force betweenqandQshould be equal to force between4qandQi.e, x2kqQ​=(l−x)2k4qQ​ ...(i)
(l−x)2=4x2
Taking square root on both sides:
l−x=2x
l=3x
x=3l​
i.e, the chargeQshould be placed at a distance3l​from chargeq.
Now, applying the condition of equilibrium on+qcharge,
⇒(3l​)2kQq​=l2k(4q)q​
⇒l2Q=4q(3l​)2
⇒Q=9×l24ql2​
⇒Q=94q​
SinceQshould be negative as explained above, soQ=9−4q​