Electrostatic force question

  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. Redbelly98

    Redbelly98 12,036
    Staff Emeritus
    Science Advisor
    Homework Helper

    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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