Deriving the equivalent capacitance in a series circuit formula When we derive the formula for the effective capacitance in series, we say:
$$Q/C_{eqv} = Q/C_1 + Q/C_2 + Q/C_3$$ (if there were 3 capacitors in this case). We would then cancel $Q$ to obtain the formula.
I understand why each capacitor has the same charge, but why does the effective capacitor have the same charge as each individual capacitor? I'd expect the effective capacitor to store a total charge of 3Q (in the given example), not Q?
When the capacitor discharges, would the overall amount of charge released not be 3Q (i.e. the overall charge of the capacitors)?
I saw a similar question on here, and it was answered by explaining that the 'inner capacitors' are isolated from the rest of the circuit, and the +Q and -Q charges cancel? But even so, the isolated charges can trigger electron flow from the 'outer capacitors' during discharge.
If anyone can clear up these doubts, I would be grateful.
 A: Just think of one c with the plattes some distance. Now put a conducting plate in the middle without touching the two outer plates. This plate will have one one side positiv charges, on the other side negativ charges. and it is now the same as two C in line. but you cannot get any of the charges on the inner plate. If you connect the outer plates only the charges of the outer plates will make a current , the interpolate will just adjust the charges going back to the other side of the plate, so it is true you will have a current from one side  of the plate to the other, but no cares going outside since in the some there were never any charges inside, the + and - were jaust separated.
A: 
I understand why each capacitor has the same charge, but why does the
effective capacitor have the same charge as each individual capacitor?
I'd expect the effective capacitor to store a total charge of 3Q (in
the given example), not Q?

It is because each capacitor has a charge $-Q$ on one plate and a charge $+Q$ on the other plate. When you connect them in series, the charges on the interior capacitors cancel. See Fig 1 of the following link:
https://courses.lumenlearning.com/physics/chapter/19-6-capacitors-in-series-and-parallel/
Hope this helps.
A: Let's imagine a series of three $0.6F$ capacitors, being charged by a $120V$ battery. From this website

Each capacitor will end up with $40V$ across it and a charge of $24C$, from $Q=CV$.
This can happen by charge (electrons) leaving one capacitor, e.g. the right plate of the left capacitor and ending up on the left plate of the second capacitor -  (similarly for the other capacitor) - but only $24C$ has flowed through the battery.
(big numbers, but it's just an example)
The battery has charged the combination with $24C$ using $120V$ and so the effective capacitance must be $0.2F$.
Also if the combination were to discharge through a resistor, only $24C$ would flow through it.  Charge cannot flow through a capacitor and the only charge flowing through the resistor would be due to electrons leaving the left plate of the left capacitor and a similar number in the wire moving onto the right plate of the right capacitor.
