Problem with intuitive explanation of charge distribution between three-plate capacitor

The situation I'm referring can be seen in the following link (ignore the numerical value, let's just call it $$V$$)

I want to find the charge, $$Q$$, on plate B. Im pretty sure I can just view it as two parallel capacitors and this results in $$Q=(C_1+C_2)V$$, where $$C_i$$ can be calculated with the usual formula for parallel capacitors (just suppose that we know the area and the two distances $$d_i$$ between $$AB$$ and $$BC$$). The problem arises when I try to intuitively picture this as charges moving from one plate to another, by means of the generator. The generator will move a charge equal to $$Q_1=C_1V$$ from plate $$A$$ to plate $$B$$ and a charge $$Q_2=C_2V$$ from plate $$C$$ to plate $$A$$. This will result in plate B having a negative charge equal to $$-Q_2-Q_1$$ and plates $$A$$ and $$C$$ having, respectively, charges $$Q_1$$ and $$Q_2$$. This just does not feel right to me, because capacitors usually have net charge $$=0$$, but if I consider capacitor $$AB$$ the net charge is $$-Q_2$$ and similarly for capacitor $$BC$$. Where is the flaw in my reasoning? Thanks a lot

Imagine Three boxes, enclosing respectively $$A$$, $$B$$ and $$C$$ (each box enclosing only one plate and having a top and a bottom side of area S).

Applying Gauss’s theorem on the Box $$Ba$$ surrounding $$A$$

$$Eab.S = Qa$$

Ditto for $$B$$ and $$C$$:

$$(Eab + Ebc).S = Qb$$

And:

$$Ebc.S = Qc$$

Since $$Eab = Ebc = V$$,

$$Qa = Qb = (1/2) Qbc$$