Question: Lets say we have a single plate that has a charge of $+Q$ on it. A plate with charge $-Q$ is infinite distance away. Will the plate with $+Q$ have a capacitance associated with it? Why or why not? I was thinking that because the distance between the plates is infinity, the capacitance is zero, but according to my TA, that is not the case. There is always some sort of "absolute" capacitance for the plate with $+Q$ (a capacitance will exist between the plate and air?). Can someone help me out in understanding this?

Answer: 
Consider the case of a conducting sphere with no counter electrode. 

The capacitance of the conducting sphere can be found with 

$$C=Q/V$$

and we know that at a distance $r$ from the centre of the sphere the potential $V$ is given by

$$V={1 \over 4\pi\epsilon\_0}{Q\over r}$$

provided that $r$ is greater than the radius of the sphere. Now if we consider the potential at the surface and subsititute into the equation above we get for a conducting sphere 

$$C= 4\pi\epsilon\_0 r$$

with no counter electrode and now $r$ is the radius of the sphere.

Now the plate that you are talking about is not a sphere, but still will have some similar, but more complicated expression for the capacitance.

Thus any conductor with a charge on it generates some potential even without a counter electrode and we can calculate the the capacitance with $Q/V$.
Both good answers. Note that the Earth has a self-capacitance by the same arguments. The Earth carries a net charge and really does have a capacitance. Remember that capacitance is defined in terms of the work that you must do to take a charge from one plate to the other. Two charged objects are involved. Even if the plates are nearly an infinite distance apart you still will have to do some work to, say, move a positive charge from the negative plate to the positive plate. The electric field due to the destination plate's positive charges will produce a repelling force that you will have to work against to move the test charge. This after all what we mean by potential difference. That is, work done to move a unit charge between two points. Very close to either charged plate the electric field only depends upon the charge density on that part of the plate and is not directly dependent upon the shape or location of the second plate. The proximity of the second plate may affect the value of this charge density. In idealized parallel plate capacitors the field happens to be uniform throughout between the plates. But this local field is always there in all geometries and results in a finite capacitance. C=Q/V. You can always calculate self-capacitance of anything by simply moving a second plate, or rock, or sphere to infinity and do the work calculation.