Why is $C=q/V$ constant for a capacitor? I get that if you have a parallel plate/spherical you can do the math and find it to be constant, but I don't see why this should be so for a general capacitor of any random shape.  In the title $q$ is charge on the capacitor, and $V$ is voltage across the capacitor.
 A: The completely general proof is a little subtle, and involves the properties of solutions to Laplace's equation.  Here's a sketch of it.
Imagine that we have two conductors of arbitrary shape.  We place a charge $+Q$ on conductor #1, and $-Q$ on conductor #2.  These charges will distribute themselves in a particular way, giving rise to surface charges $\sigma_1(\vec{r})$ and $\sigma_2(\vec{r})$, respectively.  We take the reference point for our potential ($V = 0$) to be at infinity.  When we do this, it will give rise to a potential everywhere in space;  we can call this our "reference solution" $V(\vec{r})$.  This function will satisfy Laplace's equation ($\nabla^2 V = 0$) everywhere in space;  it will also satisfying $\hat{n} \cdot \nabla V = -\sigma_1/\epsilon_0$ on the surface of conductor #1 and $\hat{n} \cdot \nabla V = -\sigma_2/\epsilon_0$ on the surface of conductor #2.
Now, suppose that we look at a new function $V'(\vec{r}) = \alpha V(\vec{r})$, where $\alpha$ is any real number.  This means that we have multiplied the potential difference between the conductors by $\alpha$.  What charge distribution on the conductors will give rise to this?  Well, on the surface of conductor #1, we have
$$
\sigma'_1 = -\epsilon_0\hat{n} \cdot \nabla V' = -\alpha \epsilon_0 \hat{n} \cdot \nabla V =\alpha \sigma_1,
$$
and similarly $\sigma'_2 = \alpha \sigma_2$.  In other words, the charge distributions are multiplied by $\alpha$ everywhere on the surface of both conductors.  In particular, this implies that the total charges on the conductors are $\pm Q' = \pm \alpha Q$;  and so the potential difference is always proportional to the amount of charge on the conductors.
You might be concerned that this is just one possible way for the charge to distribute itself on the conductors;  maybe when we double the total charge on a conductor, it gets extra-concentrated on some parts of the conductor and stays small elsewhere, instead of evenly doubling in density everywhere on the surface.  But there's a uniqueness theorem we can appeal to:

In a volume surrounded by conductors, the electric field is uniquely determined if the total charge on each conductor is given.  (The region as a whole can be bounded by another conductor, or else unbounded.)

(From Griffiths's Introduction to Electrodynamics, §3.1.6)
Since I have found a solution where net charge is multiplied by $\alpha$ on the conductors, I can then say that by the above uniqueness theorem, this is the only possible way for the charge to distribute itself on the conductors, and thus the potential difference between the conductors must also multiply by $\alpha$.
A: The idea behind the charge to potential difference ratio being constant for a capacitor is that if the charge on the capacitor is changed by a factor of $k$, then at all points of the capacitor, the charge elements will change (intuitively) proportionally, i.e. by the factor $k$. This follows from the fact that initially the forces on the charge elements were zero and when the charge is changed, the forces still remain zero (had the initial forces not been zero, changing the charge by the factor $k$ would have changed the forces by a factor $k^2$, but when the initial forces are zero, a scaling factor won't affect them as something times zero is just zero), and so the charges only scale up/down, there is no redistribution of charges. And as Michael Seifert points out in his answer, the uniqueness theorem makes sure that this possible charge distribution is the only possible charge distribution when the charge of the capacitor is changed.
I will go ahead with a rigorous proof:
Let the capacitor be composed of two conductors $C_1$ and $C_2$ of arbitrary shape, with surfaces $S_1$ and $S_2$ respectively.
If the capacitor has a charge $q_{initial}$ initially, one may write the potential of a point $P$ of $C_1$ as
$$V_P = \int_{S_1} \frac{dq}{4 \pi \epsilon_0 \ r} + \int_{S_2} \frac{dq}{4 \pi \epsilon_0 \ r}$$
($r$ being the position vector of $P$ with respect to the charge element $dq$.
And since $C_1$ is a conductor, $V_p=V_{C_1}$. Thus,
$(V_{C_1})_{initial} = \int_{S_1} \frac{dq}{4 \pi \epsilon_0 \ r} + \int_{S_2} \frac{dq}{4 \pi \epsilon_0 \ r}$.
If we now change the charge on the capacitor by a factor of $k$, then all charge elements will be changed by the same factor. Now,
$$(V_{C_1})_{final} = \int_{S_1} \frac{k \ dq}{4 \pi \epsilon_0 \ r} + \int_{S_2} \frac{k \ dq}{4 \pi \epsilon_0 \ r}  = k \ (V_{C_1})_{initial}$$
It can be similarly shown that $(V_{C_2})_{final} = k \ (V_{C_2})_{initial}$.
Observe now that 
$$\frac{q_{final}}{(V_{C_1})_{final}-(V_{C_2})_{final}} = \frac{k \ q_{initial}}{k \ (V_{C_1})_{initial}- k \ (V_{C_2})_{initial}}  = \frac{q_{initial}}{(V_{C_1})_{initial}-(V_{C_2})_{initial}}$$
Clearly, from the equation above, $\frac{q}{V}$ remains constant for the capacitor.
