Relating the Minimizing Integral to the Capacity of a Cube

[Mathmos6]

Homework Statement

The capacity C of an object is the integral over its surface
-\int_S \frac{\partial \phi}{\partial n} dA,
where the potential φ(x) satisfies Laplace’s equation in the volume outside the object, \phi = 1 on S and \phi \to 0 at \infty. Show that the capacity of a sphere of radius R is 4\pi R. (I've done that bit).

Now I need to show that the capacitance of a cube is s.t. 2 \pi a < C < 2\sqrt{3} \pi a. The hint says I need to "relate the minimizing integral (below) to the capacity. Then for the lower bound, use the volume outside the inscribing sphere and take w equal to the solution to Laplace’s equation outside the cube which is extended by w=1 in the gap between the sphere and the cube.".

Homework Equations

The 'minimising integral' is (I've proven)

\int_V |\nabla w|^2 dV \geq \int_V |\nabla u|^2 dV where u and w are both equal to f on 'S' enclosing 'V', w has continuous first partial deriv.s and u is a solution to Laplace's equation.

The Attempt at a Solution

We know \phi is going to be a function of (r) by symmetry, but I can't really even see how to begin the second part - relating the minimising integral to the capacity. I've played around with a number of identities to try and make the surface integral look like the volume one, but to no avail... help!

 

[Mathmos6]

Right, i think I've got a little further:

So you can consider \nabla \cdot (\phi \nabla \phi) = (\nabla \phi)^2 + \phi \nabla ^2 \phi, so since \phi = 1 on the relevant surfaces, \int_S \nabla \phi \cdot n dA = \int_S \phi \nabla \phi \cdot n dA? In which case by divergence theorem capacity = -\int_V \nabla \cdot (\phi \nabla \phi) dV = -(\int_V (\nabla \phi)^2 + \phi \nabla ^2 \phi dV)? At which point you'd want the integral for the volume outside the insphere = integral of (volume between insphere & cube + volume outside cube)?