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Electrostatics question

  1. Jun 20, 2005 #1
    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. [tex]\mathbf{L}_{23} = \mathbf{L}_{12}[/tex], what is the ratio of [tex]\mbox{$\frac{q_1}{q_2}}[/tex]?

    Fig. 1

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

    [tex]\mathbf{L}_{12}[/tex] is the distance between Particles 1 and 2.
    [tex]\mathbf{L}_{23}[/tex] 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:

    [tex] \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} [/tex]

    I'm not sure this will work, but I'm not sure how to accomplish this task. Any ideas?
  2. jcsd
  3. Jun 20, 2005 #2
    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.
  4. Jun 20, 2005 #3
    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.
  5. Jun 20, 2005 #4
    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.
  6. Jun 21, 2005 #5
    ok, so then:

    [tex] \Sigma\overrightarrow{F_3} = \overrightarrow{F_{13}} + \overrightarrow{F_{23}} = 0[/tex]

    [tex]\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[/tex]

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