Equipotential Field line equation derivation

[counterphit]

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

 

[eljose79]

Dear Andrew,

Thank you for your question. It sounds like you are working on a fascinating project using Solidworks and Excel to create 3D models of dipole arrangements. I am not an expert in this field, but I will do my best to explain the steps for solving the equation you have provided.

First, let's rewrite the equation to make it a bit simpler:

T = I1/r1^2 + I2/(r1^2 + 2br1 + b^2)

To solve this equation, we need to find the value of r1 that will make T equal to a given target intensity. This is known as solving for a variable. In this case, the variable we are solving for is r1.

To solve for r1, we need to isolate it on one side of the equation. Let's start by subtracting T from both sides of the equation:

T - T = I1/r1^2 + I2/(r1^2 + 2br1 + b^2) - T

On the left side, T - T equals zero, so we are left with:

0 = I1/r1^2 + I2/(r1^2 + 2br1 + b^2) - T

Next, let's combine the two fractions on the right side into a single fraction by finding a common denominator. The common denominator in this case is r1^2 + 2br1 + b^2. To add fractions with the same denominator, we simply add the numerators together and keep the denominator the same. This gives us:

0 = (I1 + I2)/ (r1^2 + 2br1 + b^2) - T

Now, let's multiply both sides of the equation by (r1^2 + 2br1 + b^2) to get rid of the fraction on the right side:

0 = (I1 + I2) - T(r1^2 + 2br1 + b^2)

We are getting closer! Next, we need to expand the parentheses on the right side of the equation by distributing T to each term inside the parentheses. This gives us:

0 = (I1 + I2) - Tr1^2 - 2Tbr1 - Tb^2

Finally, let's move all the terms with r1 to one side of the equation and all the constant terms