# Thread: Puttiing the cosine of a direction cosine in terms of Cartesian coordinates

1. ## Puttiing the cosine of a direction cosine in terms of Cartesian coordinates

Here's the problem for what it is worth.
Diagram: Let there be a point charge of -q at the coordinates (x,y) = (a, 0), a point charge of -q at (-a, 0), and a point charge of 2q at (0, L). Find the dipole moment ( $\vec{p}$) term of the electric potential and put it in terms of Cartesian coordinates.
Bear with me and I'll get to the Trigonometry part.

The dipole moment is defined as
$\vec{p} = \sum q_i \overline{r_i} = (-q)a \hat{i} + (-q)(-a) \hat{i} + (2q)L \hat{j} = 2qL \hat{j}$
(The $\overline{r_i}$ is the displacement vector from the origin to the point where the charge q sits.)

Now pick a point $\vec{r}$ in three space. Specifically $\vec{r} = r_x \hat{i} + r_y \hat{j}+r_z \hat{k}$

The dipole term of the potential is calculated by
$\phi _D = \frac{1}{4 \pi \epsilon _0} \frac{\vec{p} \cdot \vec{r}}{r^2} = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2} = \frac{2qL}{r} \cdot cos(\theta _y)$
where $cos(\theta _y)$ is the direction cosine of the angle between $\vec{p}$ and $\vec{r}$. (ie. the cosine of the angle between $\vec{r}$ and the y axis, if we make the triangle form by putting the tails of $\vec{r}$ and $\vec{r}$ at the origin.)

The question then is how to write $cos( \theta _y )$ in terms of Cartesian coordinates? I could use the Law of Cosines to find the angle if I had the length of the line between the heads of $\vec{p}$ and $\vec{r}$. I could use the Law of Sines, but I don't have any of the angles in the triangle.

Any thoughts? Thanks!

-Dan

2. ## Re: Puttiing the cosine of a direction cosine in terms of Cartesian coordinates

Originally Posted by topsquark
Here's the problem for what it is worth.
Bear with me and I'll get to the Trigonometry part.

The dipole moment is defined as
$\vec{p} = \sum q_i \overline{r_i} = (-q)a \hat{i} + (-q)(-a) \hat{i} + (2q)L \hat{j} = 2qL \hat{j}$
(The $\overline{r_i}$ is the displacement vector from the origin to the point where the charge q sits.)

Now pick a point $\vec{r}$ in three space. Specifically $\vec{r} = r_x \hat{i} + r_y \hat{j}+r_z \hat{k}$

The dipole term of the potential is calculated by
$\phi _D = \frac{1}{4 \pi \epsilon _0} \frac{\vec{p} \cdot \vec{r}}{r^2} = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2} = frac{1}{4 \pi \epsilon _0} \frac{2qL}{r^2} \cdot cos(\theta _y)$
where $cos(\theta _y)$ is the direction cosine of the angle between $\vec{p}$ and $\vec{r}$. (ie. the cosine of the angle between $\vec{r}$ and the y axis, if we make the triangle form by putting the tails of $\vec{r}$ and $\vec{r}$ at the origin.)

The question then is how to write $cos( \theta _y )$ in terms of Cartesian coordinates? I could use the Law of Cosines to find the angle if I had the length of the line between the heads of $\vec{p}$ and $\vec{r}$. I could use the Law of Sines, but I don't have any of the angles in the triangle.

Any thoughts? Thanks!

-Dan
(sobs) It was right in front of me the whole time!
$\phi _D = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2}$

Ah well. Thanks anyway!

-Dan