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Feb16-13, 07:35 PM   #1
 

Capacitance of a Fractal Surface


I understand how to calculate the capacitance of an isolated conductor, say a sphere, and I know that the associated formula involves the surface area of the conductor. Does all of this make the assumption that the surface in question is convex, or at least "non-concave"? More specifically, what if the surface is extremely complex, like a fractal where the surface area might be extremely large, but parts of the surface are directly facing other parts or even hidden behind other portions and so on. Is it still the outright surface area that matters, or do the bits and pieces of the surface have to have external normals that point out into space (if you know what I mean). Put another way, if I was to fashion an extremely convoluted surface, could I wind up with an extremely high capacitance for the volume the object is occupying? Or perhaps here's a more concrete example: What if I fashioned an object consisting of hundreds of thin metal plates all connected electrically at their centres by a rod, but with an air gap between each. A one meter cube so built could have a surface area of 500 square meters, and one might even argue that that is an "exposed" surface area, albeit primarily perpendicular to the outside world. Would this just calculate out as 6 m^2 because its basically a six sided cube, or would it in fact behave according to its, say, 500 square meter "surface"?
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