Laplace’s equation – Fourier series examples 1

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 3.12 – 3.13.

Here are a few examples of calculating the Fourier coefficients for some special cases.

Example 1. Consider the infinite slot problem with the boundary at {x=0} consisting of a conducting strip with a constant potential of {V_{0}}. In this case we get

\displaystyle   c_{n} \displaystyle  = \displaystyle  \frac{2V_{0}}{a}\int_{0}^{a}\sin\frac{n\pi y}{a}dy
\displaystyle  \displaystyle  = \displaystyle  \frac{2V_{0}}{\pi}\left(1-\cos n\pi\right)

The coefficients are thus zero for even {n} and {4V_{0}/\pi} for odd {n}:

\displaystyle  c_{n}=\begin{cases} 0 & n\mathrm{\; even}\\ \frac{4V_{0}}{\pi} & n\mathrm{\; odd}\end{cases}

The potential is thus

\displaystyle  V(x,y)=\frac{4V_{0}}{\pi}\sum_{n=1,3,5,...}^{\infty}\frac{e^{-n\pi x/a}}{n}\sin\frac{n\pi y}{a}

Example 2. Now suppose the boundary at {x=0} consists of two conducting strips, insulated from each other and from the infinite sheets. The first strip, from {y=0} to {y=a/2} has a constant potential {V_{0}} while the other strip, from {y=a/2} to {y=a} is held at potential {-V_{0}}.

Here, the coefficients {c_{n}} are given by

\displaystyle   c_{n} \displaystyle  = \displaystyle  \frac{2V_{0}}{a}\left[\int_{0}^{a/2}\sin\frac{n\pi y}{a}dy-\int_{a/2}^{a}\sin\frac{n\pi y}{a}dy\right]
\displaystyle  \displaystyle  = \displaystyle  \frac{2V_{0}}{n\pi}\left[1-2\cos\frac{n\pi}{2}+\cos n\pi\right]

If {n} is odd, this comes out to zero. If {n} is even, there are two cases. First, if {n=2,6,10,...} the term in brackets is 4. If {n=4,8,12,...} the term in brackets is zero. Thus we get

\displaystyle  c_{n}=\begin{cases} \begin{array}{c} 0\\ \frac{8V_{0}}{n\pi}\\ 0\end{array} & \begin{array}{c} n\mathrm{\; odd}\\ n=2,6,10...\\ n=4,8,12...\end{array}\end{cases}

Thus the potential is

\displaystyle   V(x,y) \displaystyle  = \displaystyle  \frac{8V_{0}}{\pi}\sum_{n=2,6,10...}^{\infty}\frac{e^{-n\pi x/a}}{n}\sin\frac{n\pi y}{a}
\displaystyle  \displaystyle  = \displaystyle  \frac{8V_{0}}{\pi}\sum_{n=0}^{\infty}\frac{e^{-(4n+2)\pi x/a}}{4n+2}\sin\frac{(4n+2)\pi y}{a}

where in the last line we’ve changed the index of summation since the non-zero terms in the first sum are just those with {n=4m+2} starting at {m=0}.

Example 3. The infinite slot with the strip at {x=0} held at potential {V_{0}} has the solution

\displaystyle  V(x,y)=\frac{4V_{0}}{\pi}\sum_{n=1,3,5,...}^{\infty}\frac{e^{-n\pi x/a}}{n}\sin\frac{n\pi y}{a}

For a conductor, the surface charge density can be found from the derivative taken normal to the surface:

\displaystyle  \sigma=-\epsilon_{0}\left.\frac{\partial V}{\partial n}\right|_{x=0}

In this case, the normal to the surface is the {x} direction, so we get

\displaystyle   \sigma \displaystyle  = \displaystyle  -\epsilon_{0}\frac{4V_{0}}{\pi}\left.\sum_{n=1,3,5,...}^{\infty}\left(-\frac{n\pi}{a}\right)\frac{e^{-n\pi x/a}}{n}\sin\frac{n\pi y}{a}\right|_{x=0}
\displaystyle  \displaystyle  = \displaystyle  \epsilon_{0}\frac{4V_{0}}{a}\sum_{n=1,3,5,...}^{\infty}\sin\frac{n\pi y}{a}

This looks fine except for the problem that the series doesn’t converge. Consider {y=a/2}. The series is then a sum of an alternating sequence of {+1} and {-1}. The original series for the potential does converge at {x=0} due to the {n} in the denominator. Not sure what the solution to this is…any comments?

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Comments

  • iman  On Tuesday, 12 June 2012 at 09:38

    first of all ..I wanted to express my gratittude.
    It takes great generousity and good heart to help expand knowledge amongst others.and physics and math are so fun!you take people to explore the joy of it.
    secondly:can I help?I study physics and math too …and I also make some good problems and I wish to help you with the blog.
    once again thank you.
    Iman
    P.S. about example 3:if we consider the slot realy infinite then it should diverge.
    you see, when we get close to a slot(or the slot is “infinite”) by gauss law we have E=zigme/2e0 and this is actually -dV/dz.where z is the perpendicular axis.so when we go away from the slot our V gets zero then negative.then – negative infinity!!!
    the problem arises from this>>we dont have an infinite slot!as we get away from it V actually becomes a function of z we cant ignore!
    infinte slot>>infinite charge>>infinte energy
    so we cant have a really infinite slot.
    and if we want to try and approximate it>>if its infinite we have symmetry so that V is only a function of z.and it is nearly:V0-zigma/2e0 z

    • Jerry  On Monday, 20 May 2013 at 15:09

      Iman ur last points are not clear to me I’d like u to explain them to me, thanks.

  • Luc  On Wednesday, 10 October 2012 at 19:57

    But the first see series you got for the voltage does converge. If you use the expression Griffiths gives you in the book (though I’m not sure how he managed to explicitly sum that) on page 337, you’ll see that it can be written as a function of x and y. It looks something like: (2Vo/pi) * tan ^-1[(sin(pi y/a)/sinh(pi y/a)]
    Then you take the derivative with respect to x of this expression and you have the surface charge density.

    • Jerry  On Thursday, 16 May 2013 at 04:17

      Luc the problem is actually this one that how does author come up with his solution so directly
      Can somebody explain this?

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