Electrostatic Forces of an Equilateral Triangle

AI Thread Summary
The discussion focuses on calculating the net electric force on a 7-nC charge located at a corner of an equilateral triangle with two other charges. The user attempts to apply Coulomb's law but struggles with the vector components and the angles between the forces. Clarification is provided that the angle between the forces acting on the 7-nC charge is 120 degrees due to the triangle's geometry. The correct approach involves using the law of cosines to find the net force instead of breaking it into components. Ultimately, the net force is estimated to be around 15.06 N.
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Homework Statement


Three point charges are located at the corners of an equilateral triangle as in the figure below. calculate the net electric force on the 7-nc charge.

Each sides are .5 m. (q1=7μc) (q2=2 μc) (q3= -4 μc)
___1
__/__\
2/____\3




Homework Equations


f=k q1 q2 / r^2


The Attempt at a Solution


F12= 8.988x10^9 (7x10^-6)(2x10^-6) /.5^2= 0.5
F13=8.988x10^9 (7x10^-6)(4x10^-6) / .5^2 = 1.006656

X
F12x=.5cos60=.25
F13x=1.006656cos-60=.503328
Rx=.25+.503328= .753328

Y
F12y=.5sin60= (sqrt of 3)/4=.4330127019
F13y=1.006656cos-60=-.4358948344
Ry=-2.882132598x10^-3

sqrt of (.753328^2+(-2.883132508x10^-3)^2)=.7534N
tan-1(-2.88...x10^-3 / .753328)= -.22°

I don't know what I'm doing wrong, but the answers I'm gettin aren't correct. help. thanks!:confused:




 
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would the angle between the two forces be 60 degrees??
 
and why don't you directly find their vector sum? why break it into components?
 
yes it's equilateral
 
the angle b/w the two forces on q=7 is 120 degrees
then
Fnet^2 = F1^2 +F2^2 -2*1/2 *F1*F2
Fnet = 15.06 (around)
 

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