Capacitance of a Fractal Surface
You said, "The concept of capacitance does not really make sense if there is only one conductor," however it was in that context that I was asking the question. I'm actually pretty comfortable and familiar with real electronic circuit capacitors and their "two plates" and how they work. But what I was asking about was situations where you have just a single object and you want to know how much charge it can hold at a given potential. For example, a VanDegraff generator with a spherical top will have some particular charge (in microcoulombs) at some particular potential (in killovolts) for some particular sphere radius. Now I'm a bit fuzzy on the following statement, but here it is: I think you pretend the other plate is another sphere of infinite radius at an infinite distance, and it becomes a math problem with some value that you take the limit as it tends to infinity or some such approach.
In practice, I believe such measurements are with respect to ground which we assume is at a neutral charge and zero potential. So when there is 350kV wrt ground on the sphere of a VanDegraff generator, we can calculate not only the "capacitance" of the sphere, but can use Q=CV to solve for the charge on the sphere. And I've seen such "solved problems" in many text books, so I know the idea of calculating the capacitance of a free object is not unusual.
Now I know traditionally capacitors were constructed as long strips of almost a sort of tinfoil and wax paper and were all rolled up to cram a huge area into a small volume, and I'm good with that, however when you do that, the surfaces of the plates remain parallel to one another (with the advantage that BOTH sides of BOTH plates are "looking" at each other thereby doubling the capacitance). If you "fractalled" them all up in a convoluted mess, but ensured that the surfaces were not allowed to cross the original boundary, would the capacitance change? i.e. the wax paper must remain stretched tight as original, but the surfaces of the plates can have ruts and textures and places where the material opens into huge "caverns" with tiny openings, etc. etc. I'm not certain how I'd model that. But I have a suspicion that it wouldn't make more "capacitance".
Let me suggest a particular configuration as an example to make my point. Lets say we have a two (identical) plate capacitor, where each plate is described as follows: Each plate is a thin rectangle, but the inside has been "hollowed out." Clearly this fact would be immaterial if that were the end of it all, and the capacitance would be calculated the same as if the plate was solid. But now, lets say that we cut a sliver out in the surface of the plates so that the hollowed interior is "accessible" and each plate can "see inside" the hollowed portion of the other through this "slit". I have a lot of trouble seeing that suddenly the capacitance will now double or triple now, even though we now have the original surface, the back of the surface, and the inside surface of the opposite side, now all "accessible" because of the slit. Clearly the surface area is four times more than the area of the facing side of the plate, and three quarters of this surface area now all "face towards" the other plate. One quarter of it is on the opposite side and everyone would agree it remains excluded from discussion just as when the plate is solid.
Now when I look at this, I tend to think that because all this new surface area is not really "line of sight" to the other plate, it doesn't really count. Its a hollow, cavernous region inside the plate with a little slit leading to the outside world.
So that's in the context of a two plate capacitor. I could be wrong, but I think it makes no difference. However, in the context of a single free object in space that we are calculating the charge on, I'm not as convinced, because it DOES represent an area that free charge can occupy, because even if its "inside", its still "on the surface" ... i.e. the surface is still a surface, even if you have to go through a little slit to reach it. But I really don't know, which is why I'm posting this ...