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# Put another way the lines const are parallel to and

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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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## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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