OK, we are speaking here of instable equilibrium, that means the sum of forces on Q2 will be zero statically in the first time instant then the system will evolve.
First of all the add charge Q3 must stay on the line where Q1 and Q2 are placed, otherwise a non compesated force will be created. We can note that, in this case, the line connecting Q1 and Q2 is y=x.
If we place the charge on the first quadrant, then Q2 will be repulsed by Q1 and attracted by Q3 then the two forces will be directed on the same way so their sum will not be zero. If instead we place the charge on the third quadrant, Q2 will be replused by Q1 and attracted by Q3 with the two forces directed in opposite direction. Then this is our case.
In order to calculate the position we have to know that the force between two charges "Qa" and "Qb" place at a distance "dab" is proportional to:
f\[\propto\]Qa*Qb/dab
2
the distance between Q1 and Q2 is d12=sqrt((0-1)
2+(0-1)
2)=sqrt(2) then f12\[\propto\]Q1*Q2/sqrt(2)
now the distance Q2 and Q3 is d23=sqrt((x-1)
2+(x-1)
2)=(x-1)*sqrt(2) then f23\[\propto\]Q2*Q3/[(x-1)*sqrt(2)]
Since we want the equilibrium we have to pose |f12|=|f23| we already find the quadrant. Thus:
Q1*Q2/2=Q2*Q3/[(x-1)
2*2]
since |Q1|=|Q2|=Q and |Q3|=4*Q
Q
2/2=4*Q
2/[(-x-1)
2*2]
solving with respect to x (we need only the positive solution) we obtain x=1 and since y=x then y=1