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Limit at infinity of an electric quadrupole

  1. Sep 7, 2008 #1
    This is for my intro physics 2 class
    1. The problem statement, all variables and given/known data
    Consider the charges Q at (-a, 0), -2Q at (0, 0) and Q at (a, 0). Such a combination of charges, with zero net charge and with zero net dipole moment, is called an electric quadrupole. a. Find the electric field along the x acis, for x > a. b. Show that, for x >> a, the electric field varies by x^-4. Find the coefficient.

    2. Relevant equations
    [tex]\vec{E} = \frac{kQ}_{r^{2}}\hat{r}[/tex]

    Here is the answer I got for part a which was correct.

    [tex] \vec{E} = kQ\left[\frac{1}_{(x+a)^{2}}} + \frac{1}_{(x-a)^{2}}} - \frac{2}_{x^{2}}}\right]\hat{j}[/tex]

    3. The attempt at a solution

    This is what I got for part b:

    [tex] \vec{E} = 2kQ\left[\frac{1}_{x^{2}}} - \frac{1}_{x^{2}}}\right]\hat{j} = 0[/tex]

    I don't know how to get the book's answer of

    [tex] \frac{6a^{2}}_{x^{4}}}\hat{j} [/tex]
     
  2. jcsd
  3. Sep 7, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Monocles,

    You do not want to set:

    [tex]
    \frac{1}{(x+a)^2} \longrightarrow \frac{1}{x^2}
    [/tex]
    because you'll lose the part you're looking for.

    Have you seen the binomial series approximation formula for [itex](1+y)^p[/itex] when y is small? You can write your quantity [itex]\frac{1}{(x+a)^2}[/itex] (and the other term, too) in that form and then use the formula to get the answer.
     
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