Ratio of charges in Coulomb's Law problem?

[podboy6]

Just got out of my E&M class lecture about Coulombs Law, I'm having trouble getting off of the ground with an electrostatics question:

Question: Three charged particles lie on the x-axis (fig. 1). Particles 1 and 2 are fixed. Particle 3 is free to move, but the electrostatic force on it from particles 1 and 2 is zero. $$\mathbf{L}_{23} = \mathbf{L}_{12}$$, what is the ratio of $$\mbox{\frac{q_1}{q_2}}$$?

Fig. 1

-----O------------O----------------O-------------x (positive x axis)

$$\mathbf{L}_{12}$$ is the distance between Particles 1 and 2.
$$\mathbf{L}_{23}$$ is the distance between Particles 2 and 3

This is from a section dealing exclusively with Coulombs law, so in this case I believe I am obligated to use it. I'm thinking of relating the forces on particle 1 to particles 2 and three in:

$$\frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}||q_{2}|}{(L_{12})^2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}||q_{2}|}{(L_{23})^2}$$

I'm not sure this will work, but I'm not sure how to accomplish this task. Any ideas?

 

[whozum]

3 is free to move, but the electrostatic force on it from particles 1 and 2 is zero
Your solution is going to revolve around this fact. Draw a force diagram for Particle 3. For the force to be 0, what has to be true about the interactions between all the particles?

edit: There was another thread with this exact problem, if you do a search I'm sure you'll find it.

 

[podboy6]

For the electrostatic force to be zero from particles 1 and 2, then shouldn't,#1 both 1 and 2 are electrically neutral and particle 3 is charged, #2 particles 1 and 2 are charged and particle 3 is neutral, or #3 all particles are neutral?

If #1 or #2 are the case, then as particle 3 moves towards 2, a charge should be induced on the neutral particle.

I guess I'm still having a hard time visualizing it and getting off the ground. And its my first day of E&M summer class.

 

[whozum]

It tells you that all the particles are charged.
The net force on particle 3 is zero.
If you have two forces acting on it, then the only way this can happen is if the forces are equal and opposite.

 

[podboy6]

ok, so then:

$$\Sigma\overrightarrow{F_3} = \overrightarrow{F_{13}} + \overrightarrow{F_{23}} = 0$$

$$\frac{1}{4 \pi \epsilon_{0}} \Large [ \normal \frac{|q_1||q_2|}{(2L)^2} + \frac{|q_1||q_2|}{L^2} \Large ]\normal = 0$$

is this sort of on the right track? I wound up getting $$q_2 = -\frac{1}{4}q_1$$ a little later on down the road.