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Calculate the electric potential inside the pipe

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    A long pipe is split to four four pieces along its length. The quarters are slightly moved apart and attached to constant electric potentials ##U_0## and ##-U_0## as shown in the cross section of the pipe in the attached picture.
    Captureh.PNG
    The walls of the pipe are thin and the distance between the quarters is small compared to radius ##a## of the pipe.
    Find the electric potential inside the pipe as function of polar coordinates ##r## and ##\varphi ## and parameters ##U_0## and ##a##. You can leave the result in infinite series.



    2. Relevant equations
    ##U(r,\varphi )=A_0+B_0ln(r)+\sum_{n=0}^{\infty}\left [ (A_mr^m+B_mr^{-m})\cos(m\varphi)+(C_mr^m+D_mr^{-m})\sin(m\varphi) \right ]##

    3. The attempt at a solution
    Now I know that all parts EXCEPT ##C_mr^m \sin( \varphi m)## should vanish in the series written in relevant equations but the problem is that I don't really understand how this happens.

    Here is how I tried:
    - Of course, we don't want the potential inside the pipe to have any infinite values at all, therefore as ##r->0## it is obvious that ##B_0=B_m=D_m=0##. This leaves me with $$U(r,\varphi )=A_0+\sum_{n=0}^{\infty}\left [ A_mr^m\cos(m\varphi)+C_mr^m\sin(m\varphi) \right ]$$

    And now I have no idea on how to continue. I have no clue what other boundary conditions should be in order to get something useful.
     
  2. jcsd
  3. Nov 14, 2014 #2

    ZetaOfThree

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    Gold Member

    The remaining boundary conditions in the problem will give you something useful. Namely, you have the potential on the inner surface of the cylinder -- use it.
     
  4. Nov 15, 2014 #3
    I thought about it but I don't get it. Can you be a bit more specific? :/
     
  5. Nov 15, 2014 #4
    Ok part b) of this problem also wants me to calculate the electric field on the symmetry plane at the angle of ##45 °##.

    Meaning ##U(r,\varphi =\frac{\pi }{4})=U(r,\frac{5\pi }{4})##. But this only leaves me with ##A_m=-C_m##. I really don't understand what my boundary conditions are in order to get rid of the ##cos##.
     
  6. Nov 15, 2014 #5
    As ZetaOfThree wrote, you can use the boundary conditions on the cylinder's surface. From the figure in your first post (if you assume that the distance between the quarters is negligible) you have ##U(a,\varphi )## (the potential at ##r=a##) for every ##\varphi \in [0,2\pi)##. It is:
    ##
    U(r=a,\varphi)=U_0\left\{\begin{matrix}
    U_0 \;\;\;,\; 0<\varphi<\pi /2\\
    -U_0\;,\; \pi/2<\varphi<\pi \\
    U_0\;\;\;,\;\; \pi <\varphi<3\pi /2\\
    -U_0\;,\; 3\pi /2<\varphi<2\pi \\
    \end{matrix}\right.##

    Taking into account the orthogonality of ##\sin ## and ##\cos ## functions, you can use the above equation to compute some integrals in order to find both ##A_m## and ##C_m## for every ##m\in \mathbb{N}## (remember Fourier's trick!). If my calculations are correct you are right that there are no ##\cos ## terms.
     
  7. Nov 15, 2014 #6
    Am... What are those integrals? Could you show one example (let's say for ##0<\varphi<\pi /2##) and than I will hopefully understand what Fourier trick is - maybe we just named it differently.
     
  8. Nov 15, 2014 #7
    Saying Fourier's trick, I mean the usual method one can use to find the coefficients (here ##A_m## and ##C_m##) in series with ##\cos m\varphi ## and ##\sin m\varphi ## (Fourier Series). For example if ##U(r,\varphi )## is given by the last equation you wrote in your first post, then it is:
    ##\displaystyle{\int _{0}^{2\pi }{U(a,\varphi )\cos n\varphi \; d\varphi }}=A_na^n##, for every ##n \in \mathbb{N}##

    Now you can compute the same integral using the equation I wrote above (the boundary conditions) for ##U(a,\varphi )## and find ##A_n##.
    In general, methods that involve integrating orthogonal functions are very common in electrostatics problems where you solve Laplace equation, so I guess that you know how to apply them. However, If you are not familiar with this method, I am not sure if this problem is suitable to apply it for the first time.
     
  9. Nov 16, 2014 #8
    I was afraid this might be it. I never understood this method but it seems like now is the time.

    So if I didn't make any mistakes than $$\int _{0}^{2\pi }{U(a,\varphi )\cos n\varphi \; d\varphi }=2\frac{U_0}{n}\sin(n\frac{\pi }{2})+2\frac{U_0}{n}\sin(n\frac{3\pi }{2})=A_na^n$$

    Now how do I see that ##A_n =0## for all ##n## ?
     
  10. Nov 16, 2014 #9
    Ah... Ok, maybe I see it now.

    Using some basic ##\sin## and ##\cos## identities $$2\frac{U_0}{n}\sin(n\frac{\pi }{2})+2\frac{U_0}{n}\sin(n\frac{3\pi }{2})=\frac{4U_0}{n}\sin(n\pi )\cos(n \cdot "something")$$

    Obviously ##\sin(n\pi )=0## for all ##n\in \mathbb{N}## meaning also that ##A_n=0## for all ##n\in \mathbb{N}##.
    This would leave me with ##U(r,\varphi )=A_0+\sum _{n=1} ^{\infty }C_nr^n\sin(n\varphi )##. Am I allowed to say that ##A_0=0## because electric potential is always defined up to a constant?
    Would this be the right conclusion?
     
  11. Nov 16, 2014 #10

    Orodruin

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    No, you can get ##A_0## out in the same way simply integrating ##U(a,\phi)## wrt ##\phi##. The constant is fixed by fixing the potential of the boundary, if you shift by a constant, you would also shift the boundary conditions.
     
  12. Nov 16, 2014 #11
    Thank you guys, I was able to get the correct result! Thank you very much!
     
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