Why Is Capacitance Not Measured in Coulombs? I understand that the simplest equation used to describe capacitance is $C = \frac{Q}{V}$. While I understand this doesn't provide a very intuitive explanation, and a more apt equation would be one that relates charge to area of the plates and distance between them, I'm having trouble understanding it in general. Capacitance seems to be describing, well, the capacity of two plates to store charge (I understand that the electric field produced between them is generally the focus more so than the actual charge). Shouldn't it just be measured in units of charge such as coulombs? I'm sure this is due to a lack of more fundamental understanding of electric potential and potential difference but I'm really not getting it.
 A: Capacitance is not a measure of how much charge that is stored on the plates.
It is a measure of how much charge that is stored per volt of sustained voltage.
Just like how, say, pressure is not the amount of force exerted, but the amount of force per area exerted. You mustn't ignore the qualifier in such definitions.
Thus, naturally, the units of capacitance are not merely Coulomb but Coulomb per volt. (This can be rewritten to Coulomb-squared-per-Joule, if you wish, since a volt is a name for Joule-per-Coulomb, a sometimes more used SI combination.) This unit combination is then given a name: Farad. The other definition of capacitance that you refer to is in terms of geometric parametres along with the permittivity - that combination ends up with equivalent units, although it is not obvious.
A: The definition of capacitance, $C=Q/V$, suggests that it should be measured in the units of charge per units of potential.
Remark: What is more amusing is that in some system of units (e.g., in cgs) the units of capacitance turn out to be the units of length.
A: An analogy here would be to a pressure vessel and asking what mass of air will fit inside.
While the tank has a fixed volume, the amount of air that will go inside depends on the pressure you that you use to force it in.  For quite a while the relationship is linear.  At double the pressure, you have double the mass of air.
Similarly, the capacitor doesn't have a fixed amount of charge that will fit.  The amount depends on the electrical "pressure" (voltage) that is used.
Actually your initial equation is the useful one.  Unless we're constructing one, we usually do not care about the physical particulars of a capacitor. Instead we want to know how much charge will move if we change the voltage.   For a "larger" capacitor (higher capacitance), more charge will fit at a given voltage.
A: Capacitance, as you describe it, is capacity to store charge - it's not charge itself. So why you expect it to be measured in unit of charge?
For SI Units, it has been decided to measure every physical quantity in terms of only 7 base units, namely, second, meter, kilogram, ampere, kelvin, mole and candela. All other units, called derived units, are to be defined using how they are related to two or more of the base units.
Since charge is current multiplied to time, its SI unit is $A.s$. Next, since electric potential is potential energy divided by charge, its SI unit becomes $kg.m^2.s^{-2}/A.s$ which is named as volt to honor Alessandro Volta. Now, since capacitance is charge divided by potential (difference), its SI unit becomes $A^2.s^4.kg^{-1}m^{-2}$ which is named as farad to honor Michael Faraday.
A: It has been found experimentally that the charge stored, $Q$ on a "capacitor" is proportional to the potential difference between its two plate, $V$.
The constant of proportionality is called the capacitance and has the unit $\rm C\,V^{-1} = F$, the farad.
Now consider your idea of defining the capacitance of a capacitor such that it has the unit the coulomb, ie it is a measure of the amount of charge stored.
The test is to see if your definition of capacitance is useful.
I select a capacitor from a drawer and put a charge of $3\,\rm C$ on it, thus according to you the capacitor has a capacitance of $3\,\rm C$.
My colleague discharges the capacitor and puts a charge of $7\,\rm C$ on it, thus according to you the capacitor has a capacitance of $7\,\rm C$.
What value label is put on the capacitor, $3\,\rm C,\, 7\,\rm C,\,  . . . . . .$?
Whilst doing the calibration I found the potential difference between the plates to be $3\,\rm V$ and my colleague found the potential difference was $7\,\rm V$.
Using the definition $C= \frac QV$ my colleague and i both agree that the capacitance of the capacitor is $\frac 3 3$ and $\frac 77$ and thus can label the capacitor as having a capacitance of $1\,\rm F$.
I would suggest that the definition $C= \dfrac QV$ is going to be more useful than your definition?
A: 
I'm sure this is due to a lack of more fundamental understanding

... but I'd say it's not about lack of understanding the concept in electricity studies, but a much, much more general concept: there are absolute quantities (like charge) and then there are relative quantities (which sometimes are derivatives), where we compare two for a good reason. This is highly general, that virtually any quantity you have, both absolute and relative can be useful and practical. Examples:

*

*I can pay a specific price for a purchase of candy, but the candy probably has a unit price -- cost per number, so a relative quantity.

*In some situation I might end up walking 2,0 km as the absolute distance, but it is simply useful to cover my inherent capability of walkinng and measure that as speed.

*In a very specific case where I use my microwave oven with speecific settings and put in 400 grams of a specific food for heating, some specific heating time is required. But in physics and generally for understanding how things work, it is useful to grasp an understanding of how hard or easy the food itself as a material is to heat up. Thus we end up dividing time by mass which lets us predict different situations and outcomes.

It sounds like you're "mentally stuck" with the absolute state of a specific, given system with a specific choice of capacitor and voltage. Comparing the capacity (as mentioned in other answers) to the voltage is a derivative concept which does not care about the voltage between its plates, but only about properties of the specific kind of capcitor. Only when you know the voltage can you then know how much charge there actually is, but that's just a completely different quantity in a very special case.
It is understandable that sometimes one can be well-versed and comfortable with using formulas but struggle understanding why they are so. It is a valid concern. In this case the answer is simply not about electromagnetism but a more general concept -- and definitions which you cannot change.
A: 
Why Is Capacitance Not Measured in Coulombs?

Because that's not how it's defined.

Capacitance seems to be describing, well, the capacity of two plates to store charge

Yes, at a certain voltage difference between the plates. That amount changes as the voltage changes.

Shouldn't it just be measured in units of charge such as coulombs?

If you specify the voltage, then you can determine the charge that would be stored, and it seems like this is your main confusion.
