# Mathemathical Methods to Solve a Physics Problem

## Homework Statement

An infinite hollow conducting cylinder of unit radius is cut into four equal parts by planes $$x=0, y=0$$. The segmments in the first and third quadrant are maintained at potentials $$+V_{0}$$ and $$-V_{0}$$ respectively, and the segments in the second and fourth quadrant are maintained at zero potential. Find $$V(x,y)$$ inside the cylinder.

## Homework Equations

This type of problem we have done with using conformal map transformations.
In the $$z-plane$$ with $$z=x+iy$$, in polar coordinates we have:

$$r=\sqrt{x^2+y^2}$$
$$\theta=\arctan{y/x}$$

## The Attempt at a Solution

In order to solve I tried conformal map transformation:
$$w=u+iv$$ with $$w=\ln{z}=\ln{x+iy}=\ln{r}+i\theta$$
In doing so then,
$$u=\ln{r}$$ and $$v=\theta$$
Using laplace equation
$$\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0$$

Similiarly Laplace equation holds true even in the $$w-plane$$. So that,
$$\frac{\partial^{2}V(x,y)}{\partial (u^2)} + \frac{\partial^{2}V(x,y)}{\partial (v^2)} =0$$

Since $$v=\theta$$ is a constant then for
$$0\leq\theta\leq\frac{\pi}{2}$$

$$V(x,y)=\frac{V_{0}}{\frac{\pi}{2}}*v=\frac{2V_{0}}{\pi}*v$$

So that converting back in the $$z-plane$$ we get:

$$V(x,y)=\frac{2V_{0}}{\pi}*\theta=\frac{2V_{0}}{\pi}*\arctan{y/x}$$

$$\frac{-\pi}{2}\leq\theta\leq\frac{-3\pi}{2}$$

$$V(x,y)=\frac{-V_{0}}{\frac{-\pi}{2}}*v=\frac{-2V_{0}}{-\pi}*v$$

$$V(x,y)=\frac{-2V_{0}}{-\pi}*\theta=\frac{2V_{0}}{\pi}*\arctan{y/x}$$

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## Answers and Replies

Can someone please comment or suggest me if this is a correct solution?

I don't have time to look at the problem in depth, but I would think that you could write the potential in terms of Bessel functions and solve for the constants using the boundary conditions. I don't know if that's right, but just a suggestion.

cazlab,
thanks for you suggestion but I think I HAVE TO solve it with conformal map transformations.

Can someone else please suggest a solution or a comment on this exisiting solution?

Can someone please suggest how to solve this problem. The above solution is not correct because apparently it assumes that the potential along the x=0 and y=0 planes is constant but it is not. It is only constant at the boundries around the circle as described above.

Here is the hint that we were given:
Find the solution of the following simpler problems: the cylinder is cut into
two equal parts by the plane y = 0, with the upper half maintained at potential +V0/2
and the lower half maintained at potential −V0/2. Use the superposition
principle to solve the original problem.

Other hint*: In order to solve the problem of Hint 1, use the following conformal transformation

$$w=\frac{i(1-z)}{1+z}$$ where $$z=x+iy$$

that maps the interior of the cylinder’s crossection onto the upper half of the w-plane.

Doing this transformation i found that

$$w=u+iv=\frac{y^2+y}{(x+1)^2+y^2}+i\frac{x^2-y^2+1}{(x+1)^2+y^2}$$

what we know is that Laplace's equation still holds for both z-plane and w-plane

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$$\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0$$
$$\frac{\partial^{2}V(u,v)}{\partial (u^2)} + \frac{\partial^{2}V(u,v)}{\partial (v^2)} =0$$