Help finding the equilibrium position of an electric field

AI Thread Summary
The discussion focuses on finding the equilibrium position of an electron in relation to two charges, q1 and q2. The user initially struggles with the vector nature of the forces and the geometric arrangement of the charges. It is established that the equilibrium position must lie along the line connecting the two charges, and the user identifies three potential regions for the electron's placement. The conversation emphasizes the relationship between the distances from the electron to the charges, specifically that the distance to q2 is sqrt(2) times that to q1. Ultimately, the user gains clarity on how to set up equations to determine the electron's position.
connor8771
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Homework Statement
A charge of -6 microCoulombs is located at (0,0). A second charge of 12 microCoulombs is located at (1,0.5). Find the x and y coordinates where an electron will be at equilibrium.
Relevant Equations
Fe=k*e*q/r^2
I seem completely lost at this. I barely know where to begin. I know that the forces will sum to 0 but the vectoral nature of the question is really confusing me. Best I have is that the distance between e and q2 has to be sqrt(2) times the distance between e and q1. I don't know where to go after this. Thank you for any help.
 
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connor8771 said:
Best I have is that the distance between e and q2 has to be sqrt(2) times the distance between e and q1.
That's a very good start. Did you draw a picture? The three points locating the 3 charges have a nice geometrical feature.
 
TSny said:
Welcome to PF!

That's a very good start. Did you draw a picture? The three points locating the 3 charges have a nice geometrical feature.
Yes, I have, but I haven't found any nice geometric feature.

If the equilibrium position(s) is along the line connecting the two charges then the force vectors will be equal and opposite, and I assume that that is the case, but I haven't found a way to prove that equilibrium positions can't exist outside of that line.
 
Suppose the electron is not on the line passing through q1 and q2. Consider the line segment connecting q1 and the electron and the line segment connecting q2 and the electron. Could these line segments be parallel with one another?
 
No, they could not, and I'm starting to see that it should have been rather easy to prove that it can't exist outside of that line.

So if they must exist along this line, then the vector from q1 to e must be of the form (x,x/2) and the vector from q2 to e must be of the form -(x,x/2). r1 must be of the form sqrt(1.25x) and r2 must be of the form sqrt(2)*sqrt(1.25x). Is this enough information to solve? I still can't really visualize what I have to do.
 
You know that the electron must be positioned on the line passing through q1 and q2. Have you decided roughly where the electron must be placed? There are three regions to consider:

(1) somewhere on the line in the third quadrant
(2) somewhere on the line between q1 and q2
(3) somewhere on the line out beyond q2

Which of these is where you need to place the electron?
 
It has to be in quadrant 3, if it was between it would simply be pulled towards the negative charge and if it was beyond q2 the repulsive force would outsize the attractive force at all distances.
 
Good. (Actually the electron would be pushed away from the negative charge since the electron is negatively charged. But I think you have the right idea.) If you knew the distance of the electron from q1, then it shouldn't be hard to determine the x and y coordinates of the position of the electron. Can you set up an equation to find the distance between the electron and q1?
 
I know the distance from q2 has to be sqrt(2) times the distance from q1 and also that the distance from q2 has to be the distance from q1 + sqrt(1.25). I think.
 
  • #10
connor8771 said:
I know the distance from q2 has to be sqrt(2) times the distance from q1 and also that the distance from q2 has to be the distance from q1 + sqrt(1.25). I think.
That's it. Just make an equation out of those words.
 
  • #11
Alright, that was a lot easier than I made it out to be. Thank you for all the help.
 
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