# 3 point electrical charge problem

1. Jun 16, 2014

### ievans_wsu

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. Jun 16, 2014

### dauto

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°

3. Jun 16, 2014

### ievans_wsu

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!