Charge stored in Series Capacitor = $\frac{Q}{3}$ I was reading a problem in class concerning capacitors of equal capacitance C in class today, and I wanted to ask a question about it. 
If we consider 3 of these capacitors connected in series:
--||--||--||--
then I know that the magnitude charge on each of these capacitors is the same.
However, if I was to plot a graph of current against time, and knowing that 
$I \times \triangle T = \triangle Q$
the book states that the total charge stored by the 3 capacitors in series is $\frac{Q}{3}$.
I'm confused by this. 
Why is this the case if each capacitor stores charge of magnitude Q, when all are connected in series?

 A: It looks like this figure is using "Q" to express the amount of charge that one capacitor would have when given a certain potential difference "V".
Now, when we hook 3 of these capacitors up in series, since they are identical the potential across them will be V/3, so the charge on each capacitor is Q/3.
Similarly, if 3 of these capacitors are hooked up in parallel, each capacitor has a potential of V, so each capacitor will have a charge Q.
Now let's think about what the graph is actually saying. We are looking at currents over time. In series, we only need to "pull" Q/3 amount of charge from the battery to get the final configuration, whereas we need to "pull" 3Q in the parallel configuration.
When we talk about "total charge" here we do not just add up the amount of charge on each plate. Adding up the associated "q" for each capacitor does not make much physical sense here, since technically when we say a charge "q" is on a capacitor what we mean is that one plate has net +q on it and the other has net -q on it. Instead, what we need to do is think about the total charge that we needed to "move" or "store". Hence this is why we have Q/3 for the series configuration and 3Q for the parallel configuration.
A: You should know the formula for capacitance in series:
$$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + ... $$
Now, if you have three capacitors connected in series, then the formula follows:
$$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} $$
Here, our values for C are all the same:
$$\frac{1}{C_T} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} $$
$$C_T = \frac{C}{3}$$
If you imagined the 3 capacitors merging into 1 capacitor of capacitance $\frac{C}{3}$, then the potential difference across it will be the same as the terminal p.d. V (facts about series circuits). 
$$C = \frac{Q}{V}$$
$$CV = Q$$
Assuming V is constant, that should mean the charge stored on the 3 capacitors in series is $\frac{1}{3}$ that of a normal capacitor with capacitance C on its own. 
What caught my eye is that the charging current is lower in capacitors series than in parallel, which is actually the question I want to ask - why is that? (I've only done IB physics to give you your main answer).
A: The capacity of the 3 capacitors in series is C/3. This is because the charge on the first plate is Q/3 and that on the last (6th) plate is -Q/3. Only this charge will flow through an external circuit connected to it such as a resistor. The fact that the other 4 plates also carry charges of $\pm Q/3$ has no impact on any outer circuit.
