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Calculating the electric field

  1. Jan 27, 2007 #1
    Here's the problem:

    Three charges are at the corners of an equilateral triangle as in the figure below:

    .........../.....\ L

    L = .06 m
    q_0 = 1.8*10^-6 C
    q_1 = 6.6*10^-6 C
    q_2 = -4.6*10^-6 C

    Calculate the electric field at the position of q_0 (the origin) due to q_1 and q_2.
    Answer is a vector in terms of i and j.

    My approach was to break down q_1 and q_2 into vector components. For example, to find the horizontal components, I did:

    E_x = k(q/x^2)

    where x is the horizontal distance from position q_0 and q is the charge. (k = 8.99*10^9)

    So, plugging this in for both q_1 and q_2, I got:

    q_1: E_x = -659266.67 N/C
    q_2: E_x = 114872.22 N/C

    I put a - on q_1 because of the direction of the vector at q_0. Adding these together I get -544394.45 for the total horizontal component. But this isn't right. Can anyone tell me where I've gone wrong in my thought process? Thanks.
    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 28, 2007 #2
    Do you know what the answer is?

    Umm... well, it's a vector, so you would need a cosine for your horizontal q_1 field at point q_0.
  4. Jan 28, 2007 #3
    It's online homework, so no I don't know what it is.

    Well since it is an equilateral triangle, all sides are the same length L and all angles are 60 degrees. I know that the q_1 charge is L/2 away from q_0 horizontally because it is equilateral (and Lcos60 = L/2). That is what would be substituted for x in the equation.
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