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

[Illeism]

Homework Statement

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.
$r^2 = x^2 + y^2$

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

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

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

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

Homework Equations

$\nabla^2 V = 0$
(no charge inside volume)

The Attempt at a Solution

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

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?

 

[Illeism]

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:

$V= const + \displaystyle\sum_{\text{even }i}{a_ir^2+b_iz^2}$

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

$V=a r^2 + b z^2 = a(x^2+y^2) + b z^2 + const$

And take the laplacian:

$\nabla^2 V= 0 = 2a + 2a + 2 b$
so
$2a=-b$

Substituting back into our equation for V, we get:

$V=a r^2 - 2a z^2 + const = a(r^2-2z^2) + const$

Now, its just a matter of relating our constants ($V_0, z_0$) 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!