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Electrostatics - Finding potential V(r,z) given hyperbolic boundry conditions.

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm trying to derive Equation (1) from the paper: http://idv.sinica.edu.tw/jwang/EP101/Paul-Trap/Winter 91 ajp demo trapping dust.pdf

    We are working with a cylindrically symmetric geometry along the z-axis.
    [itex]r^2 = x^2 + y^2[/itex]

    We have electrodes described by the hyperbolas:
    [itex]z^2 = z_0^2 + r^2/2[/itex]
    [itex]z^2 = r^2/2 - z_0^2[/itex]

    The top and bottom electrode (described by [itex]z^2 = z_0^2 + r^2/2[/itex]) are held at a potential [itex]V_0[/itex] with respect to our ground ring (described by [itex]z^2 = r^2/2 - z_0^2[/itex]) held at potential [itex]0.[/itex]

    We want to find the potential V(r,z) for the region inside the hyperbolas (the region containing the origin).

    The solution is:
    [itex]V(z,r) = V_0(\frac{1}{4z_0^2})(2z^2+r_0^2-r^2)[/itex]


    2. Relevant equations

    [itex]\nabla^2 V = 0[/itex]
    (no charge inside volume)


    3. The attempt at a solution

    So, for boundary conditions we have:
    [itex]V(z,r)=V_0[/itex] when [itex]z^2 = z_0^2 + r^2/2[/itex]
    [itex]V(z,r)=0[/itex] when [itex]z^2 = r^2/2 - z_0^2[/itex]

    I can't think of any appropriate image that would generate hyperbolic potentials, and following Jackson's section on boundary value problems (physics.bu.edu/~pankajm/LN/hankel.pdf) seems to be giving unnecessarily complex solutions, even though the final solution is quite simple.

    At this point I'm at a loss. Is there a better way to approach this problem?
     
  2. jcsd
  3. Aug 11, 2011 #2
    I found a solution, in case anyone is curious:

    The symmetry of the problem allows us to assert that we only will have even terms in our potential. So we have something of the form:

    [itex]V= const + \displaystyle\sum_{\text{even }i}{a_ir^2+b_iz^2} [/itex]

    Hoping that the lowest order term is dominant, we see what happens if we drop the other terms:

    [itex]V=a r^2 + b z^2 = a(x^2+y^2) + b z^2 + const [/itex]

    And take the laplacian:

    [itex]\nabla^2 V= 0 = 2a + 2a + 2 b [/itex]
    so
    [itex]2a=-b[/itex]

    Substituting back in to our equation for V, we get:

    [itex]V=a r^2 - 2a z^2 + const = a(r^2-2z^2) + const [/itex]

    Now, its just a matter of relating our constants ([itex]V_0, z_0[/itex]) for a specific equipotential and we're set!


    So, the actual method to generate the potential is starting from a saddle shaped potential (which allows for the trapping of particles; thus the motivation), and then figuring out what electrodes need to be to generate it... which is much easier than starting with the shape of the electrodes!
     
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