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

**topsquark** · June 22nd 2011, 12:26 PM

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

**topsquark** · June 22nd 2011, 12:43 PM

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

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