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## Homework Statement

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

## Homework Equations

This type of problem we have done with using conformal map transformations.

In the [tex]z-plane[/tex] with [tex] z=x+iy[/tex], in polar coordinates we have:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]

\theta=\arctan{y/x}

[/tex]

## The Attempt at a Solution

In order to solve I tried conformal map transformation:

[tex]w=u+iv[/tex] with [tex]w=\ln{z}=\ln{x+iy}=\ln{r}+i\theta[/tex]

In doing so then,

[tex]u=\ln{r}[/tex] and [tex]v=\theta[/tex]

Using laplace equation

[tex]\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0[/tex]

Similiarly Laplace equation holds true even in the [tex]w-plane[/tex]. So that,

[tex]\frac{\partial^{2}V(x,y)}{\partial (u^2)} + \frac{\partial^{2}V(x,y)}{\partial (v^2)} =0[/tex]

Since [tex]v=\theta[/tex] is a constant then for

[tex]0\leq\theta\leq\frac{\pi}{2}[/tex]

[tex]V(x,y)=\frac{V_{0}}{\frac{\pi}{2}}*v=\frac{2V_{0}}{\pi}*v[/tex]

So that converting back in the [tex]z-plane[/tex] we get:

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

[tex]\frac{-\pi}{2}\leq\theta\leq\frac{-3\pi}{2}[/tex]

[tex]V(x,y)=\frac{-V_{0}}{\frac{-\pi}{2}}*v=\frac{-2V_{0}}{-\pi}*v[/tex]

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

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