1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

'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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 'Capacity' in vector calculus
  1. Calculus on vectors (Replies: 14)

  2. Vector calculus (Replies: 12)

  3. Vector calculus -.- (Replies: 5)

  4. Vector calculus (Replies: 6)

  5. Vector Calculus (Replies: 1)

Loading...