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Homework Help: 'Capacity' in vector calculus

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

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

    Now I need to show that the capacitance of a cube is s.t. [itex] 2 \pi a < C < 2\sqrt{3} \pi a[/itex]. 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.".

    2. Relevant equations

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

    [itex]\int_V |\nabla w|^2 dV \geq \int_V |\nabla u|^2 dV[/itex] 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.

    3. The attempt at a solution

    We know [itex] \phi [/itex] 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!
  2. jcsd
  3. Feb 24, 2009 #2
    Right, i think I've got a little further:

    So you can consider [itex] \nabla \cdot (\phi \nabla \phi) = (\nabla \phi)^2 + \phi \nabla ^2 \phi [/itex], so since [itex] \phi = 1 [/itex] on the relevant surfaces, [itex] \int_S \nabla \phi \cdot n dA = \int_S \phi \nabla \phi \cdot n dA [/itex]? In which case by divergence theorem capacity = [itex] -\int_V \nabla \cdot (\phi \nabla \phi) dV = -(\int_V (\nabla \phi)^2 + \phi \nabla ^2 \phi dV) [/itex]? At which point you'd want the integral for the volume outside the insphere = integral of (volume between insphere & cube + volume outside cube)?
    Last edited: Feb 24, 2009
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