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Linear Quadrapole Problem

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Given the following charges, all on the x-axis:
    -q at x=-d
    +2q at x=0
    -q at x=d

    Show that the electric field at a point (x,y) = (0,r) (ie, on the y axis a distance r from the origin) is approximately:
    [tex]\frac{3qd^2}{4\pi\epsilon_{0}r^4}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    We were asked for an approximation, but it seems to me that an exact solution is relatively straightforward. The direction of the field from the positive charge would be along the y-axis. The field resulting from the negative charges would be from y=r toward those charges, but the x-components of those fields would cancel. Consequently, shouldn't the field simply be the field produced at r by the positive charge, minus the y-components of the fields produced by the negative charges? Am I missing something?

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2007 #2

    Doc Al

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    Staff: Mentor

    You're right, the exact solution is straightforward--but it's messy. Start by writing down the exact answer and showing that it reduces to that simple approximation when R >> d.
     
  4. Oct 21, 2007 #3
    Right, and for that I got [tex]\frac{2q}{4\pi\epsilon_{0}[\frac{1}{r^2}-\frac{r}{(r^2+d^2)^(\frac{3}{2})][/tex]

    I understand the concept of taking r>>d, but don't see any way with the above equation that gets either d^2 or a 3 in the numerator, or r^4 in the denominator.

    And thanks for your help!
     
  5. Oct 21, 2007 #4
    [tex]\frac{2q}{4\pi\epsilon_{0}[\frac{1}{r^2}-\frac{r}{[(r^2+d^2)^(\frac{3}{2})]][/tex]
     
  6. Oct 21, 2007 #5

    Meir Achuz

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    Science Advisor
    Homework Helper
    Gold Member

    Expand the denominators of the exact expression in the geometric series
    1/(1-x)=1+x+x^2+...
     
  7. Oct 21, 2007 #6
    Well, I'm having trouble getting the equation to show. I'll try breaking it up:

    For the field from the positive charges, I have [tex]\frac{2q}{4\pi\epsilon_0(r^2)}[/tex]
    For the negative charges, [tex]\frac{-2qr}{4\pi\epsilon_0[(r^2+d^2)]^(\frac{3}{2})}[/tex]
    Hopefully these will show
     
    Last edited by a moderator: Oct 22, 2007
  8. Oct 21, 2007 #7
    Meir's Suggestion

    Thank you - I had thought of this, but don't see how to get the expression in the proper form. My terms are {1/r^2 - r/[(r^2+d^2)^(3/2)]}. I can get this into a form of
    1-r^3/[(r^2+d^2)^(3/2)], but this is a form of 1-x, not 1/(1-x). Am I missing something obvious?
     
  9. Oct 22, 2007 #8

    Doc Al

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    Staff: Mentor

    Rewrite [tex](r^2+d^2)^{3/2}[/tex] as [tex]r^3(1+(d/r)^2)^{3/2}[/tex]

    Now you can take advantage of the approximation:
    [tex]\frac{1}{(1 + x)^a} \approx 1 - ax[/tex]
    (when x << 1)
     
    Last edited: Oct 22, 2007
  10. Oct 22, 2007 #9

    clem

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    Science Advisor

    It's easier to get the potential and then E=-dV/dx.
     
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