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General questions in E&M

  1. Mar 18, 2008 #1
    [SOLVED] General questions in E&M

    I have a few questions regarding some concepts I'd like to clear up in my upper level undergraduate E&M course. I have a test tomorrow, and my professor is big on essay questions explaining the theory behind our current material. So, I'm going to write down the concepts I'm not clear on, and then try to explain them best I can. I figure that even if nobody replies I can at least try to straighten some things out on my own by writing them down. PLEASE say something if I'm wrong or if I left something out. Here I go:

    1) For a given charge distribution, what are the physical conditions required in order for both the monopole moment and dipole moment to be zero? (i.e., if you have a quadripole or octipole, and you tried to calculate the potential of the charge distribution due to a monopole or dipole, why would it be equal to zero?)

    The only physical significance I can think of deals with the nature of the answer from multipole expansions - they are approximations. The first non-zero term in the sum dominates, and each subsequent term merely adds precision. So, if we have a quadripole (n=2), we know that the potential must fall off at a rate of 1/r^2. So, the first term in the sum MUST be equal to zero in order for this to be true:

    [tex]V(r)= \sum_{n=0}^{\infty}{\frac{1}{r'^{n+1}}\int{(r')^nP_n(cos(\theta))\roe(r')d\tau}}
    [/tex]

    Are there any other physical dependents?


    2) What conditions are necessary to allow the use of Laplace's equation instead of Poisson's equation in the determination of the electric potential?

    The only condition I can think of is that the charge in the area you are measuring must be zero...since Laplace's equation IS Poisson's equation with a charge density equal to zero. Am I missing something?
     
  2. jcsd
  3. Mar 18, 2008 #2
    I suppose the most physical thing I can say is that the charge distributions simply represent how fast the field falls away. If you have a couple charges in a dipole configuration then it will look zero pretty far away. If you have a few charges then they will cancel each other's fields even quicker, and eight charges even quicker than that!

    I mean, it's a series of inverse powers, so all the powers have to be in there somewhere. The choice of origin, your observation point, will also (somewhat obviously) effect the potential.

    For the second part I don't believe you are missing anything.
     
  4. Mar 19, 2008 #3

    Ben Niehoff

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    For Part 1:

    First, the quadrupole potential falls off as 1/r^3, not 1/r^2.

    Secondly, your explanation is a bit circular. To answer the question, think about what sort of charge distribution might have

    1. No net charge, and

    2. No net dipole moment.

    Hint: The simplest such charge distribution is an arrangement of four point charges. Hence the term "quadrupole". What can you add to a dipole to "cancel it out"?
     
  5. Mar 19, 2008 #4
    IMO the Poisson equation is more general, since it allow a distribution of the charge to be accounted for by the RHS term. When the distribution of the charge is not a concern, (e.g. when charges are concentrated at one point or when a rough mean field is wanted at a farfield), Laplace equation can be of good use.
     
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