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Suppose an electric field $E=-\nabla \psi$ where $\psi=-Q\ln r$ where $Q$ is just some constant and I have found its harmonic conjugate to be $-Q\theta+c$ where $c$ is some constant. What does it say about the field? I know that if I calculate the field directly from $E=-\nabla \psi$, I get $E=Q/r$ pointing radially outwards, but I am not sure how to interpret the harmonic conjugate found (is it even right?).

Update: I believe I have figured this out now. All the correct information follows from the Cauchy-Riemann equations!

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Your potential looks like the potential of an infinite line charge. So E does not point radially outwards but cylindrically outward. –  Revo Jan 31 '12 at 20:49
@Revo, yes, you are right, of course :) I was working in 2D. –  charge Jan 31 '12 at 20:57
@charge: It would be great if you can post your answer. –  Qmechanic Jan 31 '12 at 22:56

1 Answer 1

It gives you an E field which looks like the B field around a current carrying wire. You can think of it as being the solution from a "magnetic current" at the origin. Of course, it is not a single valued function, which has interesting implications I'm sure someone else will elaborate.

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