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Homework Help: Electrostatic Forcce

  1. Sep 3, 2010 #1
    Electrostatic Forcce :(

    1. The problem statement, all variables and given/known data
    In Fig. 21-26, particle 1 of charge q1 = +1.4 µC and particle 2 of charge q2 = -4.2 µC, are held at separation L = 12 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

    2. Relevant equations
    F = (Q1 * Q2)K/ r^2

    3. The attempt at a solution

    At first i thought i could go about this problem setting the particle to the left of the two
    i figured that the charge on Q2 > Q1 so its pull would be compensated by Q1's push
    (i made my particle three positive)
    then i set F31 = F32 so that their forces would balance out to 0
    that would make my y placemnt 0 as well
    but now...im stuck
    i did som canceling until i finally go to this:

    (r^2) Q2
    ------------ = ---------
    (0.12 +r)^2 Q1

    (r^2) / (0.12 +r)^2 = Q2/ Q1

    and i don't know wehre to go from there
    is there a way out or was i supposed to envision particle three somewhere else??
    Last edited: Sep 3, 2010
  2. jcsd
  3. Sep 3, 2010 #2
    Re: Electrostatic Forcce :(

    If you put your particle on left or right side of q1 and q2 so that it is closest to the one that pulls, you can set up two equations:

    Fpull = q3*1.4/d^2
    Fpush = q3*4.2/(d+12)^2

    when the push and pull forces are equal...
  4. Sep 3, 2010 #3
    Re: Electrostatic Forcce :(

    Thats wuht i did
    and i simplified it to get this

    (r^2) / (0.12 +r)^2 = Q2/ Q1

    i dint plug in the chrge values for Q1 and 2 yet
    but i hope u see the jist
    i pretty much started with waht u gave me, (except ur d was my r of course)
    but from the equation above i can't figure out how to solve for r
    so i was wondering if there was any other way to solve it?? :S
  5. Sep 3, 2010 #4
    Re: Electrostatic Forcce :(

    just equate them and solve the equation... here is a start:

    q3*1.4/r^2 = q3*4.2/(r+12)^2 (--divide both sides by q3)
    1.4/r^2 = 4.2/(r+12)^2 (--evaluate right side)
    1.4/r^2 = 4.2/(r^2+24r+144) (--multiply both side by (r^2)(r^2+24r+144) )
    1.4(r^2+24r+144) = 4.2r^2
    Last edited: Sep 3, 2010
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