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Unformatted text preview: ith a different potential at each plate. This time,
Laplace's equation depends only on ø and is therefore ∞øø = 0, meaning that ∞ is a linear
function of ø; that is,
∞ = Aø + B
We can now solve for A and B by knowing the potentials on each of the two plates, and
the angle between them.
The associated complex potential can be found by rewriting the above as
∞ = A Arg(z) + B
A Arg(z) is the imaginary part of ALn(z) or the real part of iALn(z). Thus ∞ is the real
part of
F(z) = iALn(z) + B
and so the associated conjugate potential is the imaginary part:
§(z) = A ln(z)
For a symmetry that depends only on ø, the complex potential F(z)=i ALn(z) + B is
just a linear function of iLn(z) (A, B real)
(D) Using Superposition: Potential due to two oppositely charged parallel wires
normal to the complex plane
The complex potential for a single wire at the origin is given by Example (B):
F(z) = ALn(z) + B
where we have B = 0 (since the potential is 0 at infinity. Also, the potential is undefined
at the origin, since the potential is infinite on the actual wire. We can obtain A by
measuring at the poten...
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 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

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