Mathemathical Methods to Solve a Physics Problem

  • #1

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]
 
Last edited:

Answers and Replies

  • #2
Can someone please comment or suggest me if this is a correct solution?
 
  • #3
9
0
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.
 
  • #4
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?
 
  • #5
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

[tex]w=\frac{i(1-z)}{1+z}[/tex] where [tex]z=x+iy[/tex]


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

Doing this transformation i found that

[tex]w=u+iv=\frac{y^2+y}{(x+1)^2+y^2}+i\frac{x^2-y^2+1}{(x+1)^2+y^2}[/tex]


what we know is that Laplace's equation still holds for both z-plane and w-plane
 
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  • #6
[tex]\frac{\partial^{2}V(x,y)}{\partial (x^2)} + \frac{\partial^{2}V(x,y)}{\partial (y^2)} =0[/tex]
[tex]\frac{\partial^{2}V(u,v)}{\partial (u^2)} + \frac{\partial^{2}V(u,v)}{\partial (v^2)} =0[/tex]
 

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