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Coulomb's Law and Electric Fields

  1. Apr 14, 2015 #1
    Examine the charge distribution shown:
    question.JPG

    b) What is the net electric field acting on charge 1?

    Attempt at the answer:

    E2 = Kq2/r2^2 = (9.0 x 10^9 Nm^2/C^2)(3.0 x 10^-5 C) / 2.0m^2

    E2 = 6.75 x 10^4 N/C

    E2 = E3 ( therefore, same procedure)

    eNet = Sqr.rt. 6.75 x 10^4 N/C^2 + 6.75 x 10^4 N/C^2 = 9.55 x 10^4 N/C

    angle = Tan^-1 = (6.75/6.75) = 45 deg.

    The total electric field acting on charge 1 is 9.55 x 10^4 N/C [N 45deg. E]

    Is this right?
     
  2. jcsd
  3. Apr 14, 2015 #2
    It looks OK.
     
  4. Apr 14, 2015 #3
    So the reason why charge 1 is not added to the equation is because is the adding of the vectors around it?
     
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