# Discrete Charge Distribution

1. Oct 6, 2014

### clb399

1. The problem statement, all variables and given/known data
Two test charges are located in the x–y plane. If q1 = -3.50 nC and is located at x = 0.00 m, y = 0.680 m and the second test charge has magnitude of q2 = 3.60 nC and is located at x = 1.00 m, y = 0.650 m, calculate the x and y components, Ex and Ey, of the electric field, , in component form at the origin, (0,0). The Coulomb Force constant is 1/(4π ε0) = 8.99 × 109 N·m2/C2.

2. Relevant equations
E=Kq/r^2

3. The attempt at a solution
So first I converted all of the given data.
q1= -3.5x10^-9 C
q2= 3.6x10^-9 C
r1=.680
r2=sqrt(1.1926)
angle=33degrees using tan^-1(.650/1)

Then I solved for E1y since it is entirely in the y direction:
E1y= K(-3.5x10^-9)/(.680)^2= -68.05

Then solved the x and y components of E2
E2x= K(3.6x10^-9)/(1.1926)^2(cos(33)) = 22.76
E2y= K(3.6x10^-9)/(1.1926)^2(sin(33))= 14.78

Then:
E1y+E2y
E2x

This is my newest answer that I've come up with but all of them have been wrong. I'm not sure what I'm missing or what's going wrong.

2. Oct 6, 2014

### Orodruin

Staff Emeritus
The electric field should be pointing the same direction as the force on a positive test charge. You can use this as a consistency check.

Is something else than the direction wrong? It would help us spot your error if you provided the answer from the solutions manual.

3. Oct 6, 2014

### clb399

I do not have a solutions manual, this is an online assignment. And what direction is wrong?

4. Oct 6, 2014

### Orodruin

Staff Emeritus
Both, the field of a negative charge should be pointing towards it and that of a positive away from it as a negative charge would attract and a positive repulse a positive test charge.

5. Oct 6, 2014

### clb399

What would the directions change? I have no idea how that helps me in the problem. I understand what you're saying, I just don't see how it applies.