Charged particle in centre of equilateral triangle

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  • #1
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Homework Statement


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


Homework Equations



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?
 

Answers and Replies

  • #2
berkeman
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Homework Statement


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


Homework Equations



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?
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:
 
  • #3
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Use rectangular coordinates to write the equations... :smile:
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.
 
  • #4
berkeman
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59,076
9,183
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.
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:
 
  • #5
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Can you express your work more in that format? :smile:
##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 :(
 
  • #6
berkeman
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59,076
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##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 :(
In the x-direction, Fx14 is zero, right? And Fx24 and Fx34 should be the same, differing only by their signs... And so on...
 
  • #7
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n the x-direction, Fx14 is zero, right? And Fx24 and Fx34 should be the same, differing only by their signs... And so on...
jess! I completely messed up there, i need to get some sleep :D.
Thank you!
 
  • #8
berkeman
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59,076
9,183
:smile:
 

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