# Potential near a conducting strip (Conformal maps)

1. Aug 25, 2017

### TSny

Yes. If the point charge is at some specified height h above the strip, then it looks like you need to numerically determine the position of the charge on the imaginary w axis. The mapping function from w to z apparently has no simple inverse.

2. Aug 25, 2017

### Spyro386

Love that new input. Never heard of those mappings but they seem like a good thing for those problems. However those are the inverses so getting to the actual mapping could still be pretty hard. Not to mention they still have the trig or hyperbolic functions which would still cause some kind of periodicity right?
Happy for the replies though, I haven't thought of many of things said :)

3. Aug 25, 2017

### TSny

Yes, the mapping is the "wrong way" and inverting it looks impossible. But, it can still be used for finding the equipotential lines in the z-plane by mapping the equipotential lines in the w-plane to the z-plane. Here is a plot of some of the equipotential lines.

4. Aug 25, 2017

### haruspex

Yes, and indeed it does seem there is no closed form inverse in this strip case.
But I just went back and reread the orginal question. Either I missed it before or later forgot that this a 3D set-up. Conformal mappings are 2D. Is it valid to map the XY plane conformally but leave the Z axis unchanged? I.e., can the wire be represented by a straight wire in the image? I would think not.
By using the reflex angles as TSny suggests, the "overlap" in the mapping should only be within the conductor. This makes it ok since all those points are at the same potential, so the ambiguity creates no conflict. But I have not checked whether this is the case.
I tried to apply it to my quadrant example. This might be invalid because instead of a "triangle with a vertex at infinity" I have two vertices at infinity. The map I got is $z^{\frac 23}$, which does indeed map the three non-conducting quadrants to a half plane.

5. Aug 25, 2017

### TSny

The conducting strip and the wire are assumed to extend to infinity in the + and - Z directions. So, the potential has no Z dependence. Laplace's equation in 3D reduces to Laplace's equation in 2D.

6. Aug 25, 2017

### haruspex

Ah, ok, more of a slab than a strip.

7. Aug 25, 2017

Yes.