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3 point electrical charge problem

  1. Jun 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Given 3 point charges as follows:
    1) a +2.5uC charge at (-0.20m , 0.15m)
    2) a -4.8uC charge at (0.50m , -0.35m)
    3) a -6.3uC charge at (-0.42m , 0.32m)

    What is the electric field at the origin (0,0)?

    2. Relevant equations

    E = (k * q) / (r^2) (To find the magnitude)

    k = 9.0e9 (N m^2)/C^2

    Etotal = Ex + Ey

    Vx = (Magnitude * Cos(θ) )
    Vy = (Magnitude * Sin(θ) )

    3. The attempt at a solution

    First, I plotted each point on a plane, and found the distance each point was from the origin using basic trig.

    Point 1 had a distance of .25m
    Point 2 had a distance of .61m
    Point 3 had a distance of .53m

    From there, I found the magnitude using the first stated equation.

    E1 = 3.6e5 N/C
    E2 = 1.2e5 N/C
    E3 = 2.0e5 N/C

    My textbook ends here saying that the addition of the vector magnitudes along the X axis should yield Ex. If i do that, I get E1 + E2 - E3 which equates out to be roughly 2.8e5. The answer in the back of the book states 2.1e5 is the correct answer. So, I did some online digging and found that the law of cosines is needed, but is not stated anywhere in the textbook sample problem. The angles I found were as followed

    1) 37°
    2) 35°
    3) 38°

    Using these angles in the equation of (Magnitude)(cos(Angle) I found the following new values

    E1 = 2.88e5
    E2 = 9.5e4
    E3 = 1.59e5

    Adding these in the manner stated before (E1 + E2 - E3) I get 2.24e5, which satisfies the X component of the problem. Now, when it comes to the Y component, I found that if you take the Sine of the angles above, and multiply them by the magnitude, it should result in the overall Y vector. The book states that -4.1e3 should be the answer, but I get nowhere near that answer. I get -1.58e5. It's been a while since I've taken physics so I'm a bit rusty. Are there any steps or equations that I'm not seeing? Thanks!
     
  2. jcsd
  3. Jun 16, 2014 #2
    You have to be more careful with the angles. In particular make sure you're getting the angle in the correct quadrant. I'm getting

    1) 180 - 36.9 = 143.1°
    2) -35.0°
    3) 180 - 37.3 = 142.7°
     
  4. Jun 16, 2014 #3
    Okay, I completely forgot about the sign changes between quadrants. My question though is why is point 3 180 - 37.3? Cosine is postive in Quadrant IV, I'm just not sure why 180 is being used to get the angle. Thanks!
     
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