# Partical equilibrium

1. Aug 27, 2009

### fball558

1. The problem statement, all variables and given/known data

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.

3. 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 im sorry) so L can be found by
sqrt(1^2 + .5^2)
then i know that at equilibrium the partical 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 im 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 im getting stuck solving for d.
please let me know if im doing this right.
thanks alot

(not sure if this is 'advanced' or not, so put in this forum)

2. Aug 27, 2009

### queenofbabes

"-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!