Calculate polarization energy over a set of cuboids

  • #1
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1

Main Question or Discussion Point

Dear Physics Forums members,

I have a research problem that involves electrostatics. My education is as a chemist, and thus I struggle to accurately represent my problem, so I thought that you guys could help me (and would be interested in the exercise).

Here is an image to summarize my problem.
PnBTzPq.png

So, in the starting point of my problem, I have my space divided into orthogonal cuboids. The cuboid at the center is particular: there is a charge distribution within it, ρ(r) and its dielectric constant is that of vacuum. All other orthogonal cuboids contain homogeneous, but not isotropic, continua. In the x direction, the dielectric constant is εxx, the component of the dielectric constant tensor ε, in y direction it is εyy, and in z direction it is εzz.
I explicitely represent three cuboids around the central cuboid in each direction, for a total of 73 - 1 = 342 cuboids. This big cuboid is immersed into a continuum that corresponds to the average value of the dielectric constant tensor ε.

What I would like to write is the equation for the polarization energy of this system (i. e. the energy change caused by replacing a conventional cuboid by a cuboid containing the charge distribution ρ(r) and with the dielectric constant of vacuum).
The system is electrostatic, i e there is no variation of the magnetic field with time. It obeys the two equations:
∇E = ρ/ε0
∇×E = 0

I know the dimension of the cuboid (the 3 lengths a, b and c from which the cuboid can be reconstructed).

I'm a bit lost and I don't know where to start.
Do some people have suggestions?

All the best!
 

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Answers and Replies

  • #2
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Do not hesitate to ask questions if this does not sound clear to you!
 
  • #3
DrDu
Science Advisor
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I fear the problem is only solvable numerically. There are very few systems for which it is possible to obtain analytical solutions of the resulting polarisation.
 
  • #4
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I know it, but my question is about how can I do it efficiently, because depending on the method you choose, your solution may be reached slower or faster to the point it's practically indistinguishable from an analytical solution.
 
  • #5
DrDu
Science Advisor
6,023
755
I suppose finite element methods are most adequate.
 
  • #6
7
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Thanks. I will think about it.
 

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