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Perturbation of a uniform electrostatic field by a dielectric cube

  1. Sep 15, 2011 #1
    Hi,

    Is there any way to analytically calculate the perturbation of a uniform electrostatic field by a dielectric cube.
    I know a solution exists for dielectric spheres but I haven't been able to come across the solution, when dealing with a cube.

    Ohh.. and I'm assuming the simplest case, where the field is parallel to the normal of one of the cube's sides.

    Thanks in advance.
     
  2. jcsd
  3. Sep 15, 2011 #2

    Vanadium 50

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    It's going to be a mess, because you won't have Legendre polynomials in your solution. It's equivalent to solving the electric field from a cube-shaped dipole, and I would attack that by starting with the electric field from a small dipole of length a, and making a cube of length L out of many aligned dipoles, and taking the limit as a/L -> 0.
     
  4. Sep 16, 2011 #3

    vanhees71

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    Instead of spherical harmonics, I'd use the eigenfunctions of the Laplace operator from the separation ansatz in Cartesian coordinates with the appropriate boundary conditions for the dielectric.
     
  5. Sep 16, 2011 #4

    Vanadium 50

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    But the variables aren't going to separate nicely in Cartesian space.
     
  6. Sep 16, 2011 #5
    Check Smythe's "Static and Dynamic Electricity". I've found the answer to many complex situations there.
     
  7. Sep 16, 2011 #6

    vanhees71

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    Of course, the Laplacian separates in Cartesian coordinates. For the Laplace Equation,

    [tex]\Delta \Phi=0.[/tex]

    you make the ansatz [itex]\Phi(\vec{x})=X(x) Y(y) Z(z)[/itex]. Then in the usual way, you can show that

    [tex]\frac{X''(x)}{X(x)}=-k_1^2, \quad \frac{Y''(y)}{Y(y)}=-k_2^2, \quad \frac{Z''(z)}{Z(z)}=k_1^2+k_2^2[/tex]

    Then you get the set of solutions

    [tex]\Phi_{k_1,k_2,\pm}(\vec{x})=\exp \left (\mathrm{i} k_1 x+\mathrm{i} k_2 y \pm \sqrt{k_1^2+k_2^2} z \right )[/tex].

    A general harmonic function can be built out of these also in the usual way. Of course, for the present problem, you have to find the restriction due to the boundary conditions imposed by the presence of the dielectric cube first and then you can linearly superpose the remaining set of functions to find the solution for your situation.
     
    Last edited: Sep 16, 2011
  8. Sep 16, 2011 #7

    Vanadium 50

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    But that's the point - that's the part that doesn't end up looking so nice. The boundary conditions are ugly, because they are all finite. With spherical or cylindrical symmetry things are much cleaner.
     
  9. Nov 22, 2011 #8
    Following the description above and solving [itex]\frac{X''}{X}=-k_1^2[/itex], I find
    [itex]X(x)=A_1e^{k_1 ix}+A_2e^{-k_1 ix}[/itex]
    similarly
    [itex]Y(y)=B_1e^{k_2 iy}+B_2e^{-k_2 iy}[/itex] and
    [itex]Z(z)=C_1e^{k_3 z}+C_2e^{-k_3 z}[/itex], as general solutions.

    So,

    [itex]\Phi(\vec{x})=
    \left[(A_1+A_2)\cos k_1x+i(A_1-A_2)\sin k_1x\right]\cdot
    \left[(B_1+B_2)\cos k_2x+i(B_1-B_2)\sin k_2x\right]\cdot
    \left[C_1e^{k_3 z}+C_2e^{-k_3 z}\right][/itex]
    rather than what is stated above by vanhees71.

    Also for the first boundary condition I have found

    [itex]\Phi(\vec{x})\rightarrow -E_0z[/itex] as any of [itex]x,y \mbox{ or } z \rightarrow \pm \infty[/itex]
    which simply states that as we move far enough away from the cube, the potential approaches that which would exist without the cube present.

    But I'm not sure how this BC would simplify the electric potential equation.
     
  10. Nov 24, 2011 #9
    Gordianus, thank you for the reference. I just went through Smyth and unfortunately there was nothing on the subject.

    Can you think of any other references which may be useful?
     
  11. Jan 14, 2012 #10
    For any unfortunate sole that may come across this post hoping for a hint as to how to approach this problem, I would ignore the above posts.

    The problem can not be solved using the Cartesian form of Laplace. The solution is attained using Green's theorem.
     
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