# Finding where an electron would be in equilibrium

• Abbie12998
Simply replace r_1 with r_2 and solve for the new x and y coordinates.Yes, that is correct. Once you have the new x and y coordinates, you can substitute r_2 into the equation to find the equilibrium point.f

## 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.

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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Welcome to the PF.

Will any equilibrium points be along the line connecting those two points? If so, why?

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

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...

Your two r’s are strictly related by the geometry. You don’t need physics to write one in terms of the other.

berkeman