# Electrostatic Forces of an Equilateral Triangle

## 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

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! ## The Attempt at a Solution

would the angle between the two forces be 60 degrees??

and why dont you directly find thier 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)