1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor expansion of an electrostatics problem

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem has six charges that are at the corners of a regular hexagon in the xy plane, each charge a distance a from the origin. I have already solved for the electric fields in the x and y direction and now am trying to apply an approximation for the field on the x axis at where x>>a. the field along the x axis is

    [itex]E_{x}= \displaystyle\sum_{n=1}^{6} \frac{x-acos\frac{k\pi}{3}}{[x^{2}-2axcos\frac{k\pi}{3}+a^{2}]^{3/2}}[/itex]

    I am supposed to use a power series in the small quantity a/x using the method of taylor series to get to

    [itex]E_{x}= \frac{1}{4\pi\epsilon_{0}}[{\frac{6q}{x^{2}}+\frac{9qa^{2}}{2x^{4}}}][/itex]

    I haven't done this level math in a long time and I am sure that it is not too difficult, but I don't know what it means in the small quantity a/x. Do I just set x>>a in the efield equation and go about my business, if so, I don't understand where the a^2 term comes from in the numerator.
     
    Last edited: Jan 30, 2013
  2. jcsd
  3. Jan 30, 2013 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You need to use [noparse] [ tex ] [ /tex ] or [ itex ] [ /itex ] [/noparse] (without the spaces). In other words, your slashes were oriented the wrong way in your closing tags. I fixed them for you. Note that you can also surround your LaTeX code with double dollar signs or double pound signs on either side as shorthand for tex and itex tags respectively.

    Taylor series expansion: let's say you have some function f(u) of some independent variable u. The Taylor series expansion of the function around some point "b" is given by $$f(u) = \sum_{n=0}^\infty \frac{1}{n!} \left[ \frac{d^n}{du^n}f(u)\right]_{u=b} (u - b)^n $$ $$ = f(b) +\left. \frac{df}{du}\right|_{u = b} (u - b) + \frac{1}{2}\left. \frac{d^2f}{du^2}\right|_{u=b}(u - b)^2 +~\textrm{H.O.T.}$$ where H.O.T. means "higher order terms." If you think about it, if you truncate the infinite series after n terms, then this amounts to finding a (n-1)th -order polynomial approximation to the function. This approximation is exactly correct at the point u = b, but it begins to diverge away from it as u moves away from b.

    In this case, your independent variable is u = a/x, and you're taking the Taylor series expansion around the point b = 0. So it's okay to neglect higher order terms, and still have a pretty accurate approximation, as long as a/x is very close to 0. (u is very close to b). That is the situation you have in this problem.
     
    Last edited: Jan 30, 2013
  4. Jan 30, 2013 #3
    Okay, so my thinking was not too far off. Now I just need to get to work. Thanks for the clarification on both the tex and the problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taylor expansion of an electrostatics problem
  1. Taylor expansion (Replies: 8)

  2. Taylor expansion (Replies: 1)

Loading...