# 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!

## Answers and Replies

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)