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.
Best Answer
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.