Calculating Electric Field Components for Discrete Charge Distribution

[clb399]

Homework Statement

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.

Homework Equations

E=Kq/r^2

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.

 

[Orodruin]

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.

 

[clb399]

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

 

[Orodruin]

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

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.

 

[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.