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Electrostatic field question help please

  1. Oct 10, 2012 #1
    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
     
  2. jcsd
  3. Oct 10, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You have it set up correctly. Somehow you switched the final numerical answers for the x and y components.
     
  4. Oct 10, 2012 #3
    OH nvm ignore that, i got it thanks stupid mistake
     
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