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Homework Help: Coloumbs Law

  1. Jul 8, 2008 #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 = +3.5 µC, x1 = 1.5 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.
    (a) Find the magnitude of the electrostatic force on q2. Done I got this part =
    94.98868N
    (b) Find the direction of this force. Got this also =
    -15.9453959° (counterclockwise from the +x axis)
    (c) At what coordinates should a third charge q3 = +5.5 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?
    Can't get this - saw problem solving methods for questions that are very similar in previous postings, but I can't get the math to work out

    2. Relevant equations
    Got as far as q1/x^2 = q2/(x+r)^2
    r being the distance between q1 and q2 since q3 must lie to the right of q1


    3. The attempt at a solution
    many attempts at a solution - but its been 15 years since I have done significant math and I can't get anywhere now.
     
  2. jcsd
  3. Jul 8, 2008 #2
    the equation you gave in part 2 is may not be right since q3 may not be located on the same Y value as q1. The equation would look more like:
    [tex]\frac{q_1q_3}{x^2+y^2} = \frac{q_2q_3}{(r+\sqrt{x^2+y^2})^2}[/tex] (where x is distance from q1 x-coordinate to q3 x-coordinate and y is dist from q1 y-coord to q3 x-coord)

    Now I'm not sure if this is the only way to solve this but I would start with equating the x-components of the forces aka [tex]F_{1-3x} = F_{2-3x}[/tex] and then do the same for y.
     
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