Particle Equilibrium Homework: Find x & y Coordinates

[fball558]

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)

 

[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!

 

[tomcue]

Your approach is mostly correct, but there are a few things that need to be clarified. First, the equation for the force should be written as F = kQq/r^2, where Q and q are the charges of the two particles and r is the distance between them. Also, since we are dealing with an electron, the charge should be -1.6 x 10^-19 C, not -1 C.

To find the distance d, you can use the Pythagorean theorem to solve for d in the equation -1/d^2 = 11/(d+L)^2. This should give you two possible solutions for d, one positive and one negative. Since we are looking for the distance from the origin, the positive solution is the correct one.

Once you have found the correct value for d, you can plug it into the equations x = -d cos theta and y = -d sin theta to find the x and y coordinates of the equilibrium position. Remember to use the correct value for theta, which in this case is arctan(0.5) = 26.6 degrees.

Overall, your approach is on the right track and just needs a few adjustments to get the correct solution. Keep up the good work!