Force equalibrium

1. Jan 16, 2006

leon1127

Three point charges are arranged along the x-axis. Charge q_1=−4.05 nC is located at x_1 = 0.195 m, and charge q_2 = 2.80 nC is at x_2=−0.340 m A positive point charge q_3 is located at the origin.Where along the x-axis can q_3 be placed and the net force on it be zero, other than the trivial answers of x= \pm \infty?
Thx

2. Jan 16, 2006

andrewchang

show some work.

how do you think it should be solved?

3. Jan 17, 2006

leon1127

[tex]\frac{4.05* q_3}{(-0.34-x)^2} =\frac{-2.8 * q_3}{(0.195+x)^2} [\tex]

I dont know why it doesnt work

Last edited: Jan 17, 2006
4. Jan 18, 2006

leon1127

may anyone?

5. Jan 18, 2006

andrewchang

draw the diagram:

|------x2-----------------------x----------------x1----------|

if x is some value, then to find the distance, you have to use the distances as (x-x2) and (x1-x)

6. Jan 18, 2006

leon1127

how do you know if x will land in the interval between [tex] q_1 and q_2 [\tex]
the problem i get is when i assume the q3 will at equalibrium, x become imaginary when i take the square root

2.8/x^2 == -4.05 / (0.535+x)^2

i have shifted the system to the right. Then the negative sign became PAIN under the square root

Last edited: Jan 18, 2006
7. Jan 18, 2006

andrewchang

crap.

i'm sorry, i didn't read the question carefully enough. as far as i can tell, it has to be either left of q_2 or right of q_1. just from eyeballing the figure, it looks like it will be to the left of q_2 because the smaller repelling charge needs to be closer to cancel out the larger attracting charge.

8. Jan 18, 2006

leon1127

nvm i got it.
Thank you everyone