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Electrostatic force question

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Three charged particles form a triangle: particle 1 with charge Q1=80nC is at xy coordinates (0,3.00mm) particle 2 with charge q2 is at (0,-3.00mm) and particle 3 is at(4.00mm,0). In unit vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) -80nC?

    The book lists the answers as: (a).(0.829N)i and (b) (-.621N).




    2. Relevant equations

    F= (|Q1||Q2|*K)/r^2

    k=permittivity constant 8.99x10^9
    r= distance between the charges

    theta= arctan(y/x)
    3. The attempt at a solution

    I have tried countless different ways to do it and can't seem to get the book's answer. I have played around with the numbers for over 3 hours and havent found anything that works. Any help would be GREATLY appreciated as I have no idea how close or far away I am.

    The last way I tried it was by drawing out the triangle. I found theta to be arctan(3/4)=36.869degrees

    F=((.64x10^24)*8.99x10^9)/(6x10^-3)^2=159.822 nC/m^2

    Fx=159.822cos(36.869)=-127.851
    Fy=159.822sin(36.869)=95.89132
    Fy= 159.822

    sqrt((159.822+95.89132)^2+(-127.851)^2))=285.8973

    I have no idea what the units are there after all that on the latest try, nor what that answer means.
     
  2. jcsd
  3. Sep 11, 2008 #2

    Redbelly98

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    What is the charge of particle #3?

    What does this have to do with Q3, the particle you are asked to calculate the force for?

    Yes.

    q1 q2 = (.64x10^24) -- Use Q3 instead of Q2. Is the exponent really +24?
    k = 8.99x10^9 -- Yes
    r = (6x10^-3) -- No. You need the distance between q1 and q3, not between q1 and q2.

    This will be correct once you calculate F correctly. Just keep track of whether these components are + or -.
     
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