The capacitance is the ratio of charge on the plates over the voltage applied.
$$C = \frac{Q}{V} \Leftrightarrow Q = C \cdot V$$
The calculation you show determines the capacitance from measured voltage and charge on the plates. You basically know the result you want and determine the size of the capacitor you need.
A larger capacitor, with a larger capacity, will hold a bigger charge at the same voltage. Doubling the area will double the capacitance (in case of a plate capacitor), so for 4 farads of capacity you get
$$Q = C \cdot V = 4 F \cdot 5 V = 20 C$$
The pysics works as follows: The voltage is a driving force, pushing electrons through the wires an onto the plates of the capacitor (or sucking them off on the positive pole), until the mutual repulsion of the electrons leads to a balance of foces. If you have a larger plate, the charge can distribute over a larger area, there is less "pileup" and therefore a smaller "pushback force". This is why, with larger plates, you get a bigger charge into your capacitor with the same voltage.
The electroscope can be considered a capacitor with capacity $C$, so it will carry a charge $Q = UC$ if we apply a voltage $U$. This means that the needle and the support strut will carry $Q$ and the case will carry the opposite charge.
The equal sign charge carried by the needle and the strut repels and equilibrates with the gravity (and constraint forces). So the amount of charge deposited on the needle and the strut determines the magnitude of the deflection of the needle (as this amount of charge determines the strength of the electrostatic force via Coulomb's law).
Note, that the capacity of the electroscope is not changed much by the movement of the needle (as the case is still far away from the needle).
So, actually, an electroscope always measures the voltage applied to its terminals, or equivalently (connected by the electroscopes capacity) the charge transferred to the electroscope, but never directly the charge carried by some capacity it is connected to.
As to your specific situation and the question about the constance of the charge: The charge on the plates of the capacitor actually change as well (although minutely, as the capacity of the electroscope will be small compared to the capacity of the capacitor). The voltage on the capacitor and the electroscope equilibrate, therefore charges are transferred to/from the electroscope until the voltages are equal (that is, the equilibrium state is given by solution of the equations: $Q_1/C_1 = Q_2/C_2$ and $Q_1 + Q_2 = Q$, where $Q$ is the original charge on the capacitor, before we connected the electroscope).
From this consideration follows as well, that an electroscope effectively measures charge, when for example charged from a plastic rod rubbed with cloth. The plastic rod has a capacity far lower than the electroscope, so nearly all charge will be transferred the the electroscope (and the voltage will drop drastically, in other words the rod is not nearly an ideal voltage source). So after collecting all the charge from the rod, you will effectively measure the amount of charge generated by the triboelectric effect.
Best Answer
The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied.
For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a breakdown region ("overloading" a capacitor, i.e. frying it probably kills off the linear relation...). Leakage current as well is not taken into account in the linearized $Q=CV$ equation.
No, $Q_{max}$ is just one parameter of a capacitor, which depends on the breakdown voltage. In regular operation, capacitors do generally not store the "maximum possible charge" they could, precisely because they follow $Q = CV$, and knowing $Q_\text{max}$ alone wouldn't provide information about the Q(V) behaviour up to that point. The capacitance tells you how much charge the thing will store if you apply a given voltage to it. $Q_\text{max}$ just tells you when it's full.