Here is a simplified approach to this question-I hope it is not too simplistic. Sorry I could not upload the mathematics and the illustrating diagram from my computer file. I need to learn how to do this, or I would appreciate if someone could leave some ideas.
Basically the approach by "MyUserIsThis" is intuitively sound. The analysis is not detailed enough to show how V depends on d (distance between the plates) at small and large d.
Imagine the two parallel plates $P_1$ and $P_2$ with finite Area $A_1$ and $A_2$, carrying eletric charges with uniform densities $D_1$ (charge $+Q_1$) and $D_2$ (charge $-Q_2$) respectively. The plates are placed on top of each other ($P_2$ above $P_1$). We assume uniform densities for simplicity. Now, choose two differential elements: $dA_1 = dx_1dy_1$ on $P_1$ at point ($x_1, y_1, 0$) and $dA_2=dx_2dy_2$ on $P_2$ at point ($x_2, y_2, d$). The potential difference between the plates is given from standard electrostatic theory, leading to the following general but ‘complicated’ double integral on the surfaces $A_1$ and $A_2$
$V(d)=-\frac{D_1D_2}{4\pi \epsilon_o} \int_{A1,A2} dA_1dA_2 \frac{1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+d^2}}$
However, for large $d$-values, due to the small size of the plates, the terms $(x_2-x_1)^2$ and $(y_2-y_1)^2$ are very small compared to $d^2$, so that the above equation reduces to this
$V(d)=-\frac{D_1D_2A_1A_2}{4\pi \epsilon_o} \frac1d= \frac{-Q_1Q_2}{4\pi \epsilon_o} \frac1d$
Therefore, the potential difference drops as $1/d$, which is equivalent to saying that for large $d$, the two plates see each other as point particles of charge $+Q_1$ and $-Q_2$, as mentioned in the previous answer. I hope this adds some clarity to the answer.
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.
Best Answer
For a parallel plate capacitor yes, the capacitance only depends on the plate area and plate separation.
This is a good example of knowing what your equations mean, and you are on the right track. The definition of capacitance we use $C=Q/U$ is true for a given capacitor with capacitance $C$, but it just gives a proportionality between the charge on the plates and the potential difference between them. If we wanted to rearrange the equation to show more of the "cause and effect" we would say $Q=CU$, meaning if we increase the potential difference between the plates, we know what charges will end up on the plates. However, I wouldn't say the capacitance "depends" on these things, since changing either $Q$ or $U$ does not actually change the capacitance, since if we change one the other changes in proportion according to $C$.
On the other hand, the equation $C=\epsilon A/d$ does show us actual dependencies. We can change the area and actually change the capacitance (How $Q$ and $U$ relate), or we change change the plate separation to actually change the capacitance. Changing one doesn't automatically change the other. In terms of the geometry, these are the only two things that matter. (We could include a dielectric between the plates, then there is more things that effect capacitance than just the geometry of the system).
Notice that by just having the equations there is no way for us to know any of this. Only when we know what the equations mean and what capacitance actually is physically can we make this sort of analysis of the equations in terms of what they tell us about the physical world.