# Coulomb's law and vectors

1. Sep 6, 2004

### bifodus

To find the magnitude of a force between two charges is very simple, but to get the direction of the force seems a little strange to me. The signs of the charges aren't included anywhere in the law, so does this mean that I literally have to think "the signs are opposite, therefore I will multiply the vector by -1 (or leave it positive, depending on my reference coordinates)"? This seems a little bit cumbersome and forced to me, and apparently not derived anywhere in the mathematics of it. Am I going about doing this the right way?

I'm quite accustomed with vectors (vector calculus and linear algebra background), but very new to e&m. Any help would be much appreciated.

2. Sep 6, 2004

### Tide

The signs are included in Coulomb's law.

3. Sep 6, 2004

### sinyud

"The signs of the charges are not included anywhere in the law". Hmmmmmmmmmmm.
I have a good book with an equation that should help you out.

F= C q1 q2 / r12^2 * (r12)/r12

Where the bolds are vectors, the regular font is scalar
and r12 =r1-r2
where r1 is the location of the q1 and
r2 is the location of q2

If I'm not too clear, I'll fix it up some more

4. Sep 6, 2004

### robphy

The "force on q1 due to q2" is (in agreement with sinyud)
$$\vec F_{on\ q_1\ due\ to\ q_2} = k\frac{q_1 q_2}{r_{12}{}^2} \hat r_{12}$$
where $\hat r_{12}$ is the unit vector at the target charge $q_1$ pointing away from the source charge $q_2$ and
$r_{12}$ is the distance to the target charge $q_1$ from the source charge $q_2$.

If the product $q_1q_2$ is positive (so they have like signs), then,
since $\hat r_{12}$ points away from $q_2$, it follows that$\vec F_{on\ q_1\ due\ to\ q_2}$ points away from $q_2$.
"$q_1$ is repelled by $q_2$."

If the product $q_1q_2$ is negative (so they have unlike signs), then,
since $\hat r_{12}$ points away from $q_2$, it follows that $\vec F_{on\ q_1\ due\ to\ q_2}$ points towards $q_2$.
"$q_1$ is attracted to $q_2$."

5. Sep 6, 2004

### bifodus

Ahh, thanks guys. For some reason I was attaching the unit vector to the equation for the magnitude of the force, which obviously removes the signs from the charges. Major brain fart.

Thanks again.