# Electric Field Help

1. Sep 3, 2008

### cowmoo32

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
$$\vec E = q1q2\hat{r} / 4\pi\varepsilon r^2$$

3. 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.

Last edited: Sep 3, 2008
2. Sep 3, 2008

### LowlyPion

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

3. Sep 3, 2008

### 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?

4. Sep 3, 2008

### tiny-tim

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 $1/ 4\pi\varepsilon$)

5. Sep 3, 2008

### LowlyPion

k is Coulomb's constant

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

Edit: Tiny-Tim beat me to it.

6. Sep 3, 2008

### LowlyPion

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

Sorry if I confused you.

7. Sep 4, 2008

### cowmoo32

Thanks for the help, guys.