- #1
clb399
- 3
- 0
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.