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First off, I apologize if this is in the wrong section of StackExchange, it's at the intersection of several different topics (mathematics, physics, and computer science), so I chose the section that seemed most relevant.

In my high school AP physics (not calculus-based) class, we are learning about electrostatics (electric fields and electromagnetic forces). For fun, I have decided to write a computer program for visualizing equipotential lines (these are lines such that all the points on the line have the same net electric potential) for two-dimensional situations.

The (simple) formula for the electric potential for a situation where there is only one charged object is: $$electric \ potential = \frac{k*Q_1}{r_1}$$ where $k$ is a constant, $Q_1$ is the charge of the charged object, and $r_1$ is the distance from the charged object.

For use in a Cartesian plane, I changed the formula slightly, to

$$electrical \ potential = \frac{k*Q_1}{\sqrt{(x_{c1} - x)^2 + (y_{c1} - y)^2}}$$

where ($x_{c1}$, $y_{c1}$) is the location of the charged object (we're assuming the object is a point here)

Thus, an equipotential line when there are $n$ charged objects is the set of all points (x,y) such that

$$C = \sum_{i = 1} ^n \frac{k*Q_i}{\sqrt{(x_{c i} - x)^2 + (y_{c i} - y)^2}} $$

where $C$ is the electric potential at every point on the equipotential line.

Here's an example of several equipotential lines for a situation where there are 3 charged objects (plotted in Grapher) enter image description here

I have tried simplifying the formula for the case where there are 3 charged objects (into something I know how to find solutions for) to no avail.

How can I solve this type of formula by hand? I'm most interested in cases where n = 3 (there are three charged objects), but if there's a more general solution to the problem, that would be awesome too.

I'm planning on implementing this in a computer program eventually, so having to do guess-and-check is fine. Details of the computer implementation are a question for later though, I'm mostly interested in understanding how to solve these types of relations.

EDIT: Let me restate the question in a better way: I am looking to solve the formula for (x,y) pairs given $C$, and the $Q_i$ and ($x_{c i}$, $y_{c i}$) for each point.

Drew
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  • There are admittedly lots of pairs, and you're bordering up on a classically hard problem http://en.wikipedia.org/wiki/Three-body_problem Unfortunately, as you've stated the question, it's not clear what you mean, and to solve the equation as it stands entirely without using any calculus, that is a rather big question. With calculus, you can take the gradient, and travel perpendicular to it to get the curves where potential is unchanging. – Chris Bonnell Jan 22 '14 at 22:27
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    @atomic Ahh, I feared that I might have to use calculus to solve the problem. Since I'm not super-comfortable with calc yet, I might just end up plugging in a bunch of points and finding which ones have a total electric potential of close to 0. – Drew Jan 22 '14 at 22:38
  • That's a fine solution and there's nothing wrong with it. Unfortunately, electromagnetic fields and multi-variable calculus are practically made for each other. – Chris Bonnell Jan 22 '14 at 22:41

1 Answers1

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Short answer: You've solved it as best as you can. Just add up the potentials as you've done.

Longer answer: If you take three (or more) arbitrary points, there's no symmetry to be exploited that results in a simpler equation. One point has radial symmetry, as the potential depends only on the distance from the charge. Two points have axial symmetry (there's an axis defined by the line joining the two points). Three points? No such luck. There may be some if you have an equilateral triangle or an isosceles triangle, but that's not the general case.

John
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