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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?

## 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?