Potential near a conducting strip (Conformal maps)

[Spyro386]

1. Homework Statement
We have an infinitely long wire with charge density $\lambda$ located at $x=0,y=h$. We also have a semi-infinite strip of a conductor $|x|<a/2, y<0$. We need to find the potential in all space. The hint is to use conformal maps.

Homework Equations

$$\nabla^2 = 0$$
(U=constant at the boundary of the conductor)
$$U = -\frac{\lambda}{2 \pi \epsilon_0} ln(r)$$

The Attempt at a Solution

The general idea is to conformal map the strip to a more simple geometry where I could use the method of image charges. I've had two attempts at doing that:

In first attempt I've simply first used the map $f(z) = z*i$ to rotate the strip and then using $g(z) = \frac{z+a/2i}{z-a/2i}(-ia/2)$ (sending -ia/2 to 0, 0 to ia/2 and ia/2 to inf) effectively strecthing the strip to the first quadrant. Finally I wanted to use $h(z) = z^2$ to map the first quadrant to upper half plane. I must've visualised $g(z)$ the wrong way as this map mapped my wire $z=ih$ inside the upper half plane.(When it was originally outside the conductor)

The second attempt started in a similar way. I use $f(z) = zi$ to rotate the strip. Then I use $g(z) = e^{\pi z/a}$ to map that to ${|z| >1 , Re(z)>0}$ Using $h(z) = z^{-2}$ I map this to the unit disc and then using $i(z) = \frac{z+i}{z-i}(-1)$ I map the disc to the upper half plane again. The problem this time was the use of the exponential fuction. It made my potential peroidic(which looking at the original problem it shouldn't be), since exponential is periodic in complex.Where did my thinking go wrong in these steps? Could I avoid the use of exponential functions and still map to a simple enough geometry to solve analitically somehow?

 

[haruspex]

It made my potential peroidic

Is it periodic within the original strip, or only if you project the solution beyond the original strip?

 

[Spyro386]

Is it periodic within the original strip, or only if you project the solution beyond the original strip?

The strip is a piece of a conductor and there is a wire next to it. Isn't potential constant within the conductor? And I need to find the potential outside the strip.

 

[haruspex]

The strip is a piece of a conductor and there is a wire next to it. Isn't potential constant within the conductor? And I need to find the potential outside the strip.

You only provided the bare outline of your method beyond all the mappings, so it is not clear to me how you got your potential function. You mentioned method of images, so maybe you meant only the contribution to the potential from the charge distribution within the conductor.
If you meant the actual potential in the conductor, I don't see how you could have got that. That it is constant would be the boundary condition you were plugging in.
Maybe if you post more details it will become clear.

 

[Spyro386]

You only provided the bare outline of your method beyond all the mappings, so it is not clear to me how you got your potential function. You mentioned method of images, so maybe you meant only the contribution to the potential from the charge distribution within the conductor.
If you meant the actual potential in the conductor, I don't see how you could have got that. That it is constant would be the boundary condition you were plugging in.
Maybe if you post more details it will become clear.

I'm calculating the potential created by both the infinitely long wire and the charges in conductor. Potential outside conductor is a harmonic function so i conformal map that area(outside conductor) to a half plane. Now my problem is an infinite conductor spanning the whole half plane and a wire next to it. I solve that with the method of image charges(another wire inside where the conductor would be).

 

[haruspex]

I'm calculating the potential created by both the infinitely long wire and the charges in conductor. Potential outside conductor is a harmonic function so i conformal map that area(outside conductor) to a half plane. Now my problem is an infinite conductor spanning the whole half plane and a wire next to it. I solve that with the method of image charges(another wire inside where the conductor would be).

Yes, I got that, but you are still not posting the details of your working beyond the proposed mapping.

Never mind. Does it help to observe that cosh maps the strip Re(z)>0, 0<Im(z)<π to the upper halfplane, reversibly? That's probably equivalent to your second method.
I see why it makes it periodic. It maps the next such strip up to the lower half plane, and so on alternating. At the same time, it maps negative x strips the other way. So when you apply the boundary condition that the potential is zero along the real line, to use the method of images, that maps back to saying the potential is zero along all the boundaries of the strips in this grid.

 

[Spyro386]

Ah alright so let me explain:

So I have the map k(z) a composition of above maps that maps strip to half plane. Let's say it maps the wire to a+ib. The resulting potential of a wire at a+ib and a condutor in the upper half plane is lambda/(4pi epsilon0)*log((x'-a)^2+(y'-b)^2)/(x'-a)^2+(y'+b)^2), where x'+iy' = k(x+iy). That is how I get the potential.

Cosh(z) seems like a nice idea, but that is still the sum of two exponents which would be periodic in complex right? So it doesn't solve the problem I have.

 

[haruspex]

which would be periodic in complex right?

Yes. But wouldn't similar issues arise with any map that is not 1-1 on the entire plane? E.g. with z², the inverse map of the zero potential boundary will consist of more than just the original zero potential boundary. And the point charge image used to represent the mapped strip maps back to multiple images.

 

[Spyro386]

Are you talking about $sqrt(z)$ having 2 solutions? Perhaps that could cause some problems but I don't use such map in my solution(I don't need an inverse to my map)

 

[haruspex]

Are you talking about $sqrt(z)$ having 2 solutions? Perhaps that could cause some problems but I don't use such map in my solution(I don't need an inverse to my map)

You perform a mapping and find a solution in the mapped plane. To apply that solution to the original problem you must invert the mapping.
I don't see how you can use the result from the mapped plane in any other way. I know you are using exp, not z², but both have this problematic feature. Indeed, this ambiguous inverse seems to be where the periodicity comes from.

I considered a simpler problem: first quadrant is a conductor, point charge q at -1+i. With the mapping z² this becomes conducting upper half plane, q at -2i, so image -q at 2i gives zero potential on the real line. But inverting the map gives an image charge -q at 1+i. That gives a zero potential on the +Im half axis, but not on +Re. Also, note that if the original problem had been a conductor covering first and third quadrants with +q/2 charges at -1+i and 1-i then the z² map would have produced the identical solution, but now the inverse map produces a sensible result: -q/2 charges at 1+i and -1-i.

I did wonder whether the trick is to take a linear combination of all the alternative inverse maps, a bit like summing linear combinations of solutions to homogeneous differential equations, but I could not make that work for this simpler problem either.

Do you have any examples of this map & image method working where the map is not 1-1 on the plane?

 

[Spyro386]

Don't have any examples :(.

I'm still not sure why we'd need to invert the map. Sure we would if we wanted to know where those image charges would have to be placed in our original problem but do we need to know that? Once we know the potential V(k(z)),we can just use that right?

Interesting point with your simpler problem. I don't have the answer to that one.

 

[haruspex]

Once we know the potential V(k(z)),we can just use that right?

I would say that is still effectively an inversion, but no matter. Suppose you apply that to my simple quadrant problem, asking what the potential is at -1-i. The map takes that to 2i, which is inside the conductor.

 

[TSny]

I'm not experienced with conformal mapping techniques. But it looks like a mapping of the "Schwarz-Christoffel" type might be appropriate. I tried it and it seems to work (i.e., it gives me equipotential lines for the strip and line charge that look right).

https://en.wikipedia.org/wiki/Schwarz–Christoffel_mapping
Note that in the example at this link, the upper half plane is mapped into the interior region of the strip. But you can modify the transformation so that it maps the upper half plane into the exterior region of the strip.

 

[Spyro386]

I would say that is still effectively an inversion, but no matter. Suppose you apply that to my simple quadrant problem, asking what the potential is at -1-i. The map takes that to 2i, which is inside the conductor.

Hmmm I see. That seems weird. The $z^2$ map still preserves harmonic functions right? If yes then maybe that's not even wrong, we just must look at the equation strictly in a mathematic way and forget about how a charge could be inside the conductor.

 

[haruspex]

I'm not experienced with conformal mapping techniques. But it looks like a mapping of the "Schwarz-Christoffel" type might be appropriate. I tried it and it seems to work (i.e., it gives me equipotential lines for the strip and line charge that look right).

https://en.wikipedia.org/wiki/Schwarz–Christoffel_mapping
Note that in the example at this link, the upper half plane is mapped into the interior region of the strip. But you can modify the transformation so that it maps the upper half plane into the exterior region of the strip.

For the strip, this gives the cosh mapping we already tried. At the link, it looks like the strip would be the non-conducting region with the rest of the plane conducting. That case is ok because in forming the image of the non-conducting region (i.e. mapping to the half plane) no overlap occurs. The solution is exactly the same as if the original layout were a planeful of non-conducting strips (going to left and right from the i axis) separated only by conducting wires.
Likewise, in the simpler problem I looked at, a conducting quadrant with an outside charge, if we swap that around to be a non-conducting quadrant with an internal charge, the z² mapping works.
In both problems, the difficulty arises when the non-conducting region is concave. The same maps don't work because they overlay regions that clearly would not have the same potential distribution.
Is there some trick in modifying the map that we're missing?

 

[TSny]

To go from concave to convex, change the "bending angles" $\beta$ and $\gamma$ from $\pi/2$ to $3\pi/2$.

 

[haruspex]

To go from concave to convex, change the "bending angles" $\beta$ and $\gamma$ from $\pi/2$ to $3\pi/2$.
View attachment 209756

Very neat!
I get a mix of functions for the inverse map. Choosing -1 and 1 as the images of the two corners and taking those corners to be at 0 and i: $z=f^{-1}(w)=K(\frac 12w\sqrt{w^2-1}-arcosh(w))$, where $K=\frac 1{\pi-\frac i{\sqrt 2}}$.
I hope that's wrong, because it makes it hard to figure out where the point charge maps to.

 

[TSny]

You're getting similar to what I got. I chose $K = i$ and got $z = \frac{1}{2}\left( w\sqrt{1-w^2} + \sin^{-1}w \right)$. The conducting strip then has the right orientation and position. The strip has a width $\pi/2$ which can easily be changed with an overall real scale factor.

Any point on the positive imaginary $w$ axis gets mapped to the positive imaginary $z$ axis. So, the point charge gets mapped to where we want.

 

[TSny]

Here's an example of how a rectangular patch in the upper half of the w-plane gets mapped to the z-plane.

 

[TSny]

I hope that's wrong, because it makes it hard to figure out where the point charge maps to.

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.

 

[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 :)

 

[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.

 

[haruspex]

However those are the inverses so getting to the actual mapping could still be pretty hard.

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.

they still have the trig or hyperbolic functions which would still cause some kind of periodicity right?

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.

 

[TSny]

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.

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.

 

[haruspex]

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 the 2D.

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

 

[TSny]

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

Yes.