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Potential at center of polyhedron = average of potential on surface

  • Thread starter camron_m21
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  • #1

Homework Statement



A closed volume is bounded by conducting surfaces that are the n sides of a regular polyhedron (n = 4,6,8,...). The n surfaces are at different potentials Vi, i = 1,2,3,...,n. Prove that the potential at the center of the polyhedron is the average of the potential on the n sides. (Jackson, 3rd ed., #2.28)

Homework Equations



Dirichlet boundary conditions: (Eqn 1)
[itex]\Phi(x) [/itex]= [itex]\frac{1}{4\pi\epsilon_{0}}[/itex][itex]\int_{V}\rho(x')[/itex]G(x,x')d3x' - [itex]\frac{1}{4\pi}\oint_{S}\Phi(x')\frac{\partial G}{\partial n'}da'[/itex]

where V is the volume enclosed by the surfaces S, and G is the Green's function.

Neumann boundary conditions: (Eqn 2)
[itex]\Phi(x) = \left\langle\Phi\right\rangle_{S} + \frac{1}{4\pi\epsilon_{0}}\int_{V}\rho(x')G(x,x')d^{3}x' + \frac{1}{4\pi}\oint_{S}\frac{\partial\Phi}{\partial n'}G(x,x')da'[/itex]

where [itex]\left\langle\Phi\right\rangle_{S}[/itex] is the average value of the potential over the surface S.

The Attempt at a Solution



I think this should be fairly easy to solve, and I have some idea of how to do it using either eqn 1 or 2. Since we know the potential on the surfaces, my initial thought was to use eqn 1, since we have Dirichlet boundary conditions. There are no charges in V, so the first integral is zero, but I'm not sure what to do with the normal derivative of the Green's function in the second derivative. We're evaluating it at the center, which we can place at the origin, so it would be G(0,x'). It would be nice if this were just unity, since then we'd have the average over the surface of the potential, which is what we want.

My other idea was to use eqn 2, which I'm not sure I can do since we're give the potential and not the field. BUT if I use that, then the first integral is zero due to no charge in V, and the second integral is zero since the potential is constant on each of the surfaces so the derivative would be zero (I think. Not sure that's true.) If all of that is valid, then we're just left with the average potential over the surface, and that's it.

My question: are either of these the right way to go, or is there something I'm missing?
 

Answers and Replies

  • #2
I think I figured it out; I reread the section in Jackson on Green's functions. I won't write out the full solution, but the general way to solve this is to use the Dirichlet formula, and to break up the surface integral into a sum over all the individual faces. Then realise that G(0,x') is the potential at a location x' of a 'unit' point charge at the origin, with all the surfaces at zero potential, and thus its normal derivative gives the induced charge density. From there the integral is pretty easy to evaluate. Hopefully this helps someone.
 
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