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Electric field at the origin

  1. May 14, 2006 #1
    "Three charges, +2.5 microC, -4.8microC, and -6.3microC, are located at (-0.20m, 0.15m), (0.50m, -0.35m), and (-0.42m, -0.32m) respectively. What is the electric field at the origin?"
    Using the coordinates, I found that r12 = 0.0625 m2, r22 = 0.3725m2, and r32 = 0.2788m2. I then plugged that into the equation E = (kq)/r2, where k = 9 * 10^9.

    Using that I found:
    E1 = 3.6 * 10^5 N/C at 37º below the +x-axis.
    E2 = 1.2 * 10^5 N/C at 35º above the -x-axis.
    E3 = 2.0 * 10^5 N/C at 37º below the -x-axis.
    Then the final components of the total Electric Field is:
    Ex = 2.9 * 10^4 N/C
    Ey = -2.7 * 10^5 N/C

    However, the answer is 2.2 * 10^5 N/C for x, and -4.1 * 10^5 N/C for y. What am I doing wrong?
     
  2. jcsd
  3. May 15, 2006 #2

    siddharth

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    Gold Member

    You problem solving approach looks correct.

    Maybe you went wrong in the arithemtic (+/- signs)? Or perphaps you've not calculated the angles correctly?
     
  4. May 15, 2006 #3
    It was the angles.
    E2 should be = 1.2 * 10^5 N/C at 35º below the +x-axis.
     
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