Finding where an electron would be in equilibrium

  • Thread starter Abbie12998
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Homework Statement


A point charge of -1.0 µC is located at the origin. A second point charge of 16 µC is at x = 1 m, y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium.

Homework Equations


F=k((q1q2)/r^2)

The Attempt at a Solution


I have gotten pretty far with the problem, but I have reached a block. I used the equation F=k((q1q2)/r^2) for both the point at the origin(F1) and the point at (1,0.5) (F2).
I have come up with the equations:
F1=k(-1μC)/(r_2)^2
F2=k(16μC)/(r_2)^2
Then I set them equal to each other:
((-1μC)/(r_1)^2)=((16μC)/(r_2)^2)
I know from here I should find a way to substitute r_2 to put it in terms of r_1 but I'm not sure how to go about it. I drew out where the points would be graphically, and I put the first point at the origin and the second point at (1,0.5), is that where I went wrong? Any help would be appreciated.
 
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Answers and Replies

  • #2
berkeman
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Welcome to the PF. :smile:

Will any equilibrium points be along the line connecting those two points? If so, why?
 
  • #3
Yes, it would be somewhere along the line that connects the two points because the negative charge would push the electron towards the positive charge. I just don't understand how to mathematically find the point that it reaches equilibrium. I know you would set the two forces equal to each other, but I'm not sure how to proceed
 
  • #4
berkeman
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Yes, it would be somewhere along the line that connects the two points because the negative charge would push the electron towards the positive charge. I just don't understand how to mathematically find the point that it reaches equilibrium. I know you would set the two forces equal to each other, but I'm not sure how to proceed
Set the two forces equal and opposite in a vector sense, along that line...
 
  • #5
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Your two r’s are strictly related by the geometry. You don’t need physics to write one in terms of the other.
 
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