Charged particle in centre of equilateral triangle

[prehisto]

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?

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?

 

[berkeman]

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?

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...

 

[prehisto]

Use rectangular coordinates to write the equations...

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.

 

[berkeman]

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?

 

[prehisto]

Can you express your work more in that format?

$Fx=Fx_{14} +Fx_{24}+Fx_{34}=F_{14}/cos(60)+F_{24}/cos(60)+Fx_{34}= 18+18-9=27kN$
$Fy=Fy_{14} +Fy_{24}+Fy_{34}=F_{14}*cos(90)+F_{24}*cos(30)+F_{34}*cos(30)= 0+7,24+7,24=15,48kN$
looking at the final values,looks like something is terribly wrong , but i don't see it :(

 

[berkeman]

$Fx=Fx_{14} +Fx_{24}+Fx_{34}=F_{14}/cos(60)+F_{24}/cos(60)+Fx_{34}= 18+18-9=27kN$
$Fy=Fy_{14} +Fy_{24}+Fy_{34}=F_{14}*cos(90)+F_{24}*cos(30)+F_{34}*cos(30)= 0+7,24+7,24=15,48kN$
looking at the final values,looks like something is terribly wrong , but i don't 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...

 

[prehisto]

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!

 

[berkeman]