Conductor completely surrounds another conductor A problem in my book asks to
"Demonstrate that the capacitance of any conductor is always smaller than or equal to the capacitance of a conductor which can completely surround it".
The solution to this is obvious to me for conductors which are concentric and have poloidal/azimuthal symmetry because all of the integrals involved are then very easy. However, if one has irregularly shaped conductors, is there some generalization of the integrations that can be used to show this result? 
 A: That's interesting because the proof needs a relationship between Gauss's Law, which involves a surface internal of $\vec{E}$, and a definition of potential difference, which involves a line integral of $\vec{E}$. 
If you consider just the original conductor with a positive charge, and place a Gaussian surface at the position of the outer conductor, the average electric field component perpendicular to the surface (spatially averaged over the surface) is pointing outward. Some point in the surface must actually have that value. Since the electric field points toward lower potential, the inside has higher potential. Repeat for a progression of surfaces that interpolate the space between the two conductors. That establishes that the inner surface is at higher potential than the order surface. 
Work the Gaussian surfaces in the other direction (toward infinity) and you can establish that if either surface is given the same positive charge, the outer one will be at a lower potential. Therefore the outer one has more capacitance. 
