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Equipotential Field line equation derivation

  1. Apr 17, 2013 #1
    Hello,

    This is not a homework question! I am simply trying to understand for my own amusement. A while back I synched Solidworks and Excel, and am using this to create 3d models of different things such as dipoles for visual interpretation.

    So in doing this I needed to find potential field lines of two point particles in dipole arrangement. At any point where the two radii of the particles in question plus the distance between them form a triangle, the problem was easy. However when the point in question (target) lies in line with both (as the area of the triangle approaches zero) trig is no longer an option. now it is the difference between two inverse square functions.
    [itex]\frac{I1}{r1^2}[/itex] and [itex]\frac{I2}{r2^2}[/itex]

    being, initial intensity 1 over r1 squared, and initial intensity two over r2 squared.



    I am looking for a target intensity (T) giving me the equation

    T=[itex]\frac{I1}{r1^2}[/itex]+[itex]\frac{I2}{r2^2}[/itex]

    the distance between the two particles (B for base of triangle) has the folowing relationship with the two radii

    B=r2-r1 or in other words, the line segment B plus the line segment r1 equal r2

    Therefore we can subsittute r2=r1+b into the previous equation to solve for r1


    T=[itex]\frac{I1}{r1^2}[/itex]+[itex]\frac{I2}{(r1+b)^2}[/itex]

    which factors out to



    T=[itex]\frac{I1}{r1^2}[/itex]+[itex]\frac{I2}{r1^2 + 2br1+b^2}[/itex]

    Known:

    I1 - initial intensity 1
    I2 - initial intensity 2
    B - base, or distance between point particles
    T - target intensity

    Now I have had the damnedest time trying to solve this equation in a single shot. I have not worked on math in years and am having a fantastic time learning from scratch once again. If someone would be so kind as to walk me through how to solve this problem i would be forever grateful. I know I can take both initial intensities and set them to the same potential with reversed sign, but I would like to keep the formula flexible. I would love to know what to do with the meeting of two quadratic equations, so the final answer would be nice, but the logical progression of steps would be above and beyond!

    Thanks a ton in advance!

    Andrew
     
  2. jcsd
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