Particle Equilibrium Homework: Find x & y Coordinates

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A point charge of -1 C at the origin and a charge of 11 C at (1 m, 0.5 m) create a scenario where an electron's equilibrium position needs to be determined. The user has set up the forces acting on the electron and derived an equation relating the distances involved, but is struggling to solve for the distance 'd'. A correction is noted that L should be sqrt(1.25) instead of sqrt(125). The next steps involve cross-multiplying, expanding the equation, and solving the resulting quadratic equation for 'd' to find the equilibrium coordinates. The discussion emphasizes the importance of accurate calculations in physics problems.
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Homework Statement



A point charge of -1 C is located at the origin. A second point charge of 11 C 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.



The Attempt at a Solution



i got pretty far on this (i think) just stuck on the last step. here is what i have done so far.

first i said L is an imaginary line connecting the two particles. you can form a triangle out of this and L is the hypotonuse (spelled wrong I am sorry) so L can be found by
sqrt(1^2 + .5^2)
then i know that at equilibrium the particle will have an attraction and a repulsion force (called F1 and F2)
|F1| = k*-1*e/d^2
|F2| = k*11*e/(d+L)^2
i set these equal to each other
k*-1*e/d^2 = k*11*e/(d+L)^2
and get

-1/d^2 = 11/(d+L)^2 where L = sqrt(125)

i also found that theta = arctan .5

this is where I am stuck.

i think i have to find d from the above equation and plug into
x= -d cos theta
y = -d sin theta
to get my x and y cordinate of equilibrium.
but I am getting stuck solving for d.
please let me know if I am doing this right.
thanks alot

(not sure if this is 'advanced' or not, so put in this forum)
 
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"-1/d^2 = 11/(d+L)^2 where L = sqrt(125)"

L should be sqrt(1.25)...
Simply cross multiply, expand the brackets, and solve the quadratic equation!
 
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