Calculating Electric Field Due to Multiple Charged Balls

[cowmoo32]

Homework Statement

At a particular moment, three small charged balls, one negative and two positive,are located as shown in Figure 13.46. Q1 = 3 nC, Q2 = 7 nC, and Q3 = -6 nC.

What is the electric field at the location of Q1, due to Q2?

Homework Equations

\vec E = q1q2\hat{r} / 4\pi\varepsilon r^2

The Attempt at a Solution

r = <0, .04, 0> m
rmag = .04
rhat = <0, 1, 0>

E = [3e-9(7e-9)(9e9)] / .0016

Which gives me:

E = <0, 1.18125e-4, 0>

What am I doing wrong here? At first I figured I missed a negative, so I tried making my answer negative, but apparently that's not the case.

 

[LowlyPion]

\vec E = q1q2\hat{r} / 4\pi\varepsilon r^2

The Attempt at a Solution

r = <0, .04, 0> m
rmag = .04
rhat = <0, 1, 0>

E = [3e-9(7e-9)(9e9)] / .0016

Which gives me:

E = <0, 1.18125e-4, 0>

What am I doing wrong here? At first I figured I missed a negative, so I tried making my answer negative, but apparently that's not the case.

Isn't the electric field at Q1 due to Q2 given by
E = kQ/r² = kQ2/(.04)² ?

 

[cowmoo32]

The way we've been doing it in class is the way I posted. We haven't even used k...what does it represent?

 

[tiny-tim]

\vec E = q1q2\hat{r} / 4\pi\varepsilon r^2

Hi cowmoo32!

No, that's Coulomb's law for the force …

the field is just q2\hat{r} / 4\pi\varepsilon r^2

(I think LowlyPion is saying the same thing, and his k is your 1/ 4\pi\varepsilon)

 

[LowlyPion]

The way we've been doing it in class is the way I posted. We haven't even used k...what does it represent?

k is Coulomb's constant

k = \frac{1}{4 \pi \epsilon_0}

Edit: Tiny-Tim beat me to it.

 

[LowlyPion]

I wrote it as k because Tex is more awkward to pound out.

Sorry if I confused you.

 

[cowmoo32]

Thanks for the help, guys.