Calculating Electric Field for 2 Test Charges

[jprforester]

Homework Statement

Two test charges are located in the x–y plane. If q1 = -2.75 nC and is located at x = 0.00 m, y = 0.800 m and the second test charge has magnitude of q2 = 3.20 nC and is located at x = 1.00 m, y = 0.400 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/(4*pi*ε) = 8.99 × 10^9 N·m^2/C^2

given:
q1 = -2.75 nC; x = 0.00 m, y = 0.800 m
q2 = 3.20 nC; x = 1.00 m, y = 0.400 m

Homework Equations

E=kq/r^2

The Attempt at a Solution

So what I did was take the q1 and q2 values (given in nC) and convert them to C.
Next I took found the distance from the origin of the two points, solved using the E=kq/r^2 for each point where k=8.99e+9.
I found the components of the E1 (has only j component) and E2 (by solving the angle from the origin using tan^-1(y-distance/x-distance) and multiplying that by the E2 value)
then I added the two vectors together but still do not get the right answer.

I Calculated:
q1 = -2.75e-9 C
q2 = 3.20e-9 C
r1 = .8
r2 = sqrt(1^2 + .4^2) = 1.077
theta = tan^-1(.4/1) = 21.8 deg

I solved the equation
so E1 = -38.6 j (N/C)
and |E2| = 24.8 N/C => separate into components
E2 = 23.03 i + 9.21 j (N/C)

thus Ex = 23.03 N/C
Ey = -15.6 N/C

 

[cepheid]

Welcome to PF,

Basically, the directions of your components are wrong.

In vector form, the electric field of a charge q is $$\mathbf{E} = \frac{kq}{r^2}\mathbf{\hat{r}}$$Boldface quantities represent vectors, and the unit vector $\mathbf{\hat{r}}$ is a unit vector that points "radially outward" (i.e. away from the charge that is the source of the field). The vector $-\mathbf{\hat{r}}$ points towards the charge that is the source of the field. Unlike the Cartesian unit vectors $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$, and $\mathbf{\hat{k}}$, whose directions are fixed, the direction of $\mathbf{\hat{r}}$ varies depending on where you are in space. In the case where you're located at the origin, and the source of the field, q1, is located on the y-axis at (0,0.8), the radial unit vector, which points away from the source charge and towards the location where you are evaluating the field, is in the $-\mathbf{\hat{j}}$ direction (i.e. $\mathbf{\hat{r}}$ = $-\mathbf{\hat{j}}$ in this particular case). HOWEVER because q1 is negative, this negative sign cancels out the one on the $\mathbf{\hat{j}}$, and the E-field ends up pointing in the positive $\mathbf{\hat{j}}$ direction. So E1 is directed upwards along the y-axis, towards the charge q1. This make sense, because q1 is negative, so a positive test charge placed at the origin would be attracted up towards it.

Using a similar argument, you can reason that the x and y components of E2 should be in the $-\mathbf{\hat{i}}$ and $-\mathbf{\hat{j}}$ directions respectively (again, because $\mathbf{\hat{r}}$ points away from q2 and towards the origin, and q2 is positive this time).