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Some confusion regarding yet another Coulumb's Law problem.

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data

    The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.0 microCoulombs, x1 = 3.5cm, y1 = 0.50cm, and q2 = -4.0 microCoulombs, x2 = 2.0cm, y2 = 1.5cm. Find the (a) magnitude and (b) direction of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge q3 = +4.0 microCoulombs be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?

    2. Relevant equations

    Herein lies the problem. Other than kq1q2/r^2 (Coulomb's Law), I'm not sure how to approach this.

    3. The attempt at a solution

    I've tried the method I remember from previous Physics, which is to add x1 + x2 for a net X and y1 + y2 for a net Y, then taking the square root of Xnet^2 + Ynet^2 for the magnitude, and the arctangent of Ynet over Xnet for the angle, which get me some answers, but I'm not sure if this is the proper method to use for parts a and b. I am also unclear on how to even begin parts c and d other than the observation that F31 + F32 must equal zero. Any help or hints at how to get started would be appreciated.
     
  2. jcsd
  3. May 13, 2007 #2

    hage567

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    Homework Helper

    If would be much more helpful if you could elaborate on what you've done. How did you set up your equations? What answers did you get? I'm going to guess that you've got the right idea, about breaking up the forces into the x and y components.
     
  4. May 14, 2007 #3
    No. You cant just add x1 and x2. The r^2 refers to the magnitude of the distance between them, do you know the distance formula? [tex]\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]. That is r.
     
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