(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

E=kq/r^2

3. 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

[itex]\sqrt{Ex^2+Ey^2}[/itex] = 8.99*10^13

which isnt the answer i dont know where ive gone wrong

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

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# Homework Help: Electrostatic field question help please

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