Particle Equilibrium Homework: Find x & y Coordinates

  • Thread starter Thread starter fball558
  • Start date Start date
  • Tags Tags
    Equilibrium
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
fball558
Messages
143
Reaction score
0

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 a lot

(not sure if this is 'advanced' or not, so put in this forum)
 
Physics news on Phys.org
"-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!