Three charges -q, Q and -q are placed at equal distances on . . .

Question : Three charges -q, Q and -q are placed at equal distances on a straight line. If the potential energy of the system of these charges is zero, then what is the ratio Q:q?

Doubt by Pratham 

Solution : 

A             B            C
._______._______.

-q           Q            -q

AB=BC=r 
AC=2r

Potential Energy of the system is zero. (Given)

We know, potential energy due to system of two point charges is given by 
U = kq₁q₂/r

So 

U_AB+U_BC+U_AC=0
k(-q)(Q)/r + k(Q)(-q)/r + k(-q)(-q)/2r = 0
k/r[-qQ-qQ+q²/2]=0
kq/r[-Q-Q+q/2]=0
-2Q+q/2=0
(-4Q+q)/2=0
-4Q+q=0
q=4Q
1/4=Q/q
Hence, Q:q=1:4