Puttiing the cosine of a direction cosine in terms of Cartesian coordinates

Here's the problem for what it is worth.

Quote:

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 ($\displaystyle \vec{p}$) 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

$\displaystyle \vec{p} = \sum q_i \overline{r_i} = (-q)a \hat{i} + (-q)(-a) \hat{i} + (2q)L \hat{j} = 2qL \hat{j}$

(The $\displaystyle \overline{r_i}$ is the displacement vector from the origin to the point where the charge q sits.)

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

The dipole term of the potential is calculated by

$\displaystyle \phi _D = \frac{1}{4 \pi \epsilon _0} \frac{\vec{p} \cdot \vec{r}}{r^2} = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2} = \frac{2qL}{r} \cdot cos(\theta _y)$

where $\displaystyle cos(\theta _y)$ is the direction cosine of the angle between $\displaystyle \vec{p}$ and $\displaystyle \vec{r}$. (ie. the cosine of the angle between $\displaystyle \vec{r}$ and the y axis, if we make the triangle form by putting the tails of $\displaystyle \vec{r}$ and $\displaystyle \vec{r}$ at the origin.)

The question then is how to write $\displaystyle cos( \theta _y )$ 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 $\displaystyle \vec{p}$ and $\displaystyle \vec{r}$. I could use the Law of Sines, but I don't have any of the angles in the triangle.

Any thoughts? Thanks!

-Dan

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

Quote:

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

$\displaystyle \vec{p} = \sum q_i \overline{r_i} = (-q)a \hat{i} + (-q)(-a) \hat{i} + (2q)L \hat{j} = 2qL \hat{j}$

(The $\displaystyle \overline{r_i}$ is the displacement vector from the origin to the point where the charge q sits.)

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

The dipole term of the potential is calculated by

$\displaystyle \phi _D = \frac{1}{4 \pi \epsilon _0} \frac{\vec{p} \cdot \vec{r}}{r^2} = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2} = frac{1}{4 \pi \epsilon _0} \frac{2qL}{r^2} \cdot cos(\theta _y)$

where $\displaystyle cos(\theta _y)$ is the direction cosine of the angle between $\displaystyle \vec{p}$ and $\displaystyle \vec{r}$. (ie. the cosine of the angle between $\displaystyle \vec{r}$ and the y axis, if we make the triangle form by putting the tails of $\displaystyle \vec{r}$ and $\displaystyle \vec{r}$ at the origin.)

The question then is how to write $\displaystyle cos( \theta _y )$ 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 $\displaystyle \vec{p}$ and $\displaystyle \vec{r}$. 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!

$\displaystyle \phi _D = \frac{1}{4 \pi \epsilon _0} \frac{2qLr_y}{r^2}$

Ah well. Thanks anyway!

-Dan