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Electric Field of Two Charges

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Your Open QuestionShow me another »
    Electric field help, please?
    Find the electric field at P in the figure shown below. (Take r = 1.5 m and θ = 41°. Measure the angle counterclockwise from the positive x-axis.)
    http://i834.photobucket.com/albums/zz264/kpw0629/1-1.gif


    2. Relevant equations

    E=(8.99e9*q)/r^2

    3. The attempt at a solution
    Ive been doing (8.99e9*q)/r^2=a and then multiplying a by sin of 41 degrees=b. Then, multiplying b by 2=.58 because theyre the same force and you add them together, but that's not right and I don't know what to do differently. I feel like the angle would also be 41 (or 319) because they're even forces but I don't know.
    Thanks for any help.
    (8.99e9*1/9e-9)/1.5^2=.4439*sin41=.29*2=.58
     
    Last edited: Jan 28, 2010
  2. jcsd
  3. Jan 29, 2010 #2

    Lok

    User Avatar

    sin(a)=opposite side/hypotenuse

    cos(a)= close side/hypotenuse
    which one should be used here?

    The 2 charges create the same electric field so the two add up but at an angle. Consider the 2 fields as forces acting on a point and see how they add up.
     
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