About parallel plate conductors' capacitance Capacitance of the parallel plate conductors is $εA/d$ where $A$ is the area of each plate and $d$ is the distance between them. My question is, instead of 2 identical plates, what if we used 2 identical smooth surfaces such as $sinx=z$. Would the formula for the capacitance change?
 A: In general the formula does change. For example, the capacitance of two long co-axial cylindrical conducting shells, of length $l$ and radii $a$ and $b$ ($b>a$) is
$$C=\frac {2\pi l \epsilon_0}{\ln (b/a)}$$
This cannot generally be cast into the parallel plate form, $C=\tfrac{\epsilon_0 A}d$, even though the separation of the conductors is the same everywhere.
However if $b$ is only a little larger than $a$, that is $b= a+d$ in which $d<<a$, then, using just the first term in the Maclaurin expansion of $\ln (1 + x)$,
$$C=\frac {2\pi l \epsilon_0}{\ln (b/a)}=\frac {2\pi l \epsilon_0}{\ln (1+\tfrac da)}\approx\frac {2\pi a l \epsilon_0}d,$$
which is the familiar parallel plate capacitor formula, as $2\pi a l$ is the approximate plate area !
This illustrates the first sentence of robphy's second paragraph.
A: In the parallel plate capacitor, that formula assumes infinite-sized plates.
But for real finite-size plates that are close together, it's a good approximation if you aren't near the edge of the plates.
If the separation between the plates is small and the surfaces don't wiggle too rapidly, you could try to model the curved plates as a collection of parallel-plate capacitors in parallel. (For example, if the plates were not parallel... but one had an slight incline with respect to the other.)
However, if the model fails, then you'll have to resort to a more general definition of capacitance.
