How Is the Electric Field Calculated at the Third Vertex of a Triangle?

[skg94]

Homework Statement

What is the electric field at the third vertex.

http://tinypic.com/r/350toif/6


edit- not working actual link works = http://tinypic.com/r/350toif/6

Homework Equations

E=kq/r^2

The Attempt at a Solution

E1= (labeled on the diagram) = (8.99*10^9)(4)/(.02)^2 = 8.99*10^13
E2 = same, since equal triangle

E1x= 8.99*10^13cos60=7.78556838*10^13
E1y= " sin60 = 4.495*10^13

Now if my theory is right, this x and y are n of e so both are positive.

E2x= -7.78556838*10^13
E2y= 4.495*10^13

Ex= E1x+E2x=0
Ey=E1y+E2y=8.99*10^13
$\sqrt{Ex^2+Ey^2}$ = 8.99*10^13

which isn't the answer i don't know where I've gone wrong

answers on the picture, 1.56*10^14 N/c

 

[TSny]

E1x= 8.99*10^13cos60=7.78556838*10^13
E1y= " sin60 = 4.495*10^13

You have it set up correctly. Somehow you switched the final numerical answers for the x and y components.

 

[skg94]

You have it set up correctly. Somehow you switched the final numerical answers for the x and y components.

OH nvm ignore that, i got it thanks stupid mistake