What Is the Charge of q if the Net Force on the 1.0 mC Charge Is Zero?

[RyanBruceX]

Homework Statement

In this diagram, the net force on the 1.0 mC charge is zero. What is the sign and magnitude of the unknown charge q? Please see attached photo.
[/B]

Homework Equations

F = KQ1Q2/r^2[/B]

The Attempt at a Solution

I have approached this by first determining the force required by “q” on the middle particle to maintain the system in its current position. Because the 2 bottom particles cancel in the “x” direction the combined force they place on the middle particle is 2Fy? And so “q” must also place a force of 2Fy but in the -y direction on the middle particle. So by this rationale I should be able to calculate the charge of “q” using q = (2Fy)(r^2)/(k)(Q1) where Q1 is the middle particle.
q = (8.65 x 10^6 N)(0.0009 m^2) / (9.0 x 10^9 N M2/C2)(1.0 x 10^-3 C) = 8.65 x 10^-4 C positive charge?

I am assuming I do not have to use the vector method to determine the force of q because it also has equal components in both the - and + x directions which cancel. So net force is just in the y direction and this is used to calculate the charge?

The magnitude of this charge seems low to me, I would have expected a more symmetrical charge. Should I be considering the forces between q and the bottom two particles?[/B]

 

[DrClaude]

Misread the question. Please ignore my previous post.

 

[gneill]

Your work looks fine. Note that q doesn't just produce "equal components in both the - and + x directions which cancel", it produces NO components in the in those directions when it acts on the 1mC charge (or, technically, they are zero valued).

The charge may seem small but keep in mind it is closer to the 1mC charge than the others and the force varies as the inverse square of the separation.

 

[RyanBruceX]

Thank you very much! I forgot about this being a "fixed position" question. My thoughts on the -/+ x direction components for "q" were based on the forces between "q" and the bottom particles. Would they not also produce a force at an angle towards "q" as they do towards the middle particle? I just have a hard time conceptualizing this problem, my solution was approached by breaking the system up. Focusing on the effect of the bottom particles on the middle particle only, and then focusing on the equal and opposite effect of "q" on the middle particle (forgetting fixation). So while I may have come up with the right answer it was wrong reasoning. When I look at the system as a whole It would seem to me that all the particles should exert a force on each other and so "q" should feel a force of 2Fy (using q and 2 mC as charges, and the distance q to the bottom particle) plus the Force of the middle particle on q. Unless somehow 2Fy using 1 mC and 2 mC and the distance from the bottom to middle provides the same magnitude...