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Charged particle in centre of equilateral triangle

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    3 charged particles located at equilateral triangles corners has charge ##q_1=q_2=q_2=4*10^{-6} C ## The 4th particle with charge ## q_4=3*10^{-4}## is placed at bisectors crossing. The distance between q1 and q4 is 3.46 cm .What is the net force on q4?
    trijsturis.png

    2. Relevant equations

    3. The attempt at a solution
    So I calculated the forces ## F_{43} =F_{42}=F_{41}= (k q_1 q_2 )/ r^2 =9kN ##
    Next I thought I calculate the F* and sum of ## F_{41} and F* ##
    ##F*=F^*_{43} +F^*_{42}=F_{43}/cos(60)+F_{42}/cos(60)= 36kN##

    Now when i add up ##F* +F_{41}=27kN## Its getting confusing , because I was certain that sum of all forces on q4 should be zero,because of the symmetry and q1=q2=q3.

    My question to you guys is, is my calculation wrong or the sum of all forces should not be zero?
     
  2. jcsd
  3. May 25, 2016 #2

    berkeman

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    Staff: Mentor

    I think the sum should be zero, but you need to show the vector sum of the vector forces to show it. Use rectangular coordinates to write the equations... :smile:
     
  4. May 25, 2016 #3
    Thank you for responding.

    But i think I am did exactly that. If we imagine that direction from q4 to q1 ir positive axis direction . I projected two vectors to x axis and added them up, then subtracted the third.
     
  5. May 25, 2016 #4

    berkeman

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    I guess the way you wrote it is not intuitive for me, and doesn't appear to show the forces in both the x and y directions. I was looking for something more like this:

    ΣFx = Fx14 + Fx24 + Fx34

    ΣFy = Fy14 + Fy24 + Fy34

    Can you express your work more in that format? :smile:
     
  6. May 25, 2016 #5
    ##Fx=Fx_{14} +Fx_{24}+Fx_{34}=F_{14}/cos(60)+F_{24}/cos(60)+Fx_{34}= 18+18-9=27kN##
    ##Fy=Fy_{14} +Fy_{24}+Fy_{34}=F_{14}*cos(90)+F_{24}*cos(30)+F_{34}*cos(30)= 0+7,24+7,24=15,48kN##
    looking at the final values,looks like something is terribly wrong , but i dont see it :(
     
  7. May 25, 2016 #6

    berkeman

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    Staff: Mentor

    In the x-direction, Fx14 is zero, right? And Fx24 and Fx34 should be the same, differing only by their signs... And so on...
     
  8. May 25, 2016 #7
    jess! I completely messed up there, i need to get some sleep :D.
    Thank you!
     
  9. May 25, 2016 #8

    berkeman

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    Staff: Mentor

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