Results 1 to 2 of 2

Math Help - Puttiing the cosine of a direction cosine in terms of Cartesian coordinates

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    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
    Last edited by topsquark; June 22nd 2011 at 12:46 PM. Reason: Fixed the Physics
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1

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

    Quote Originally Posted by topsquark View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. secant inverse in terms of cosine inverse?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2010, 08:08 PM
  2. Why is cosine(-x) equal to cosine(x)?
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: August 6th 2010, 05:56 PM
  3. Lowering Powers - Simplify in terms of cosine
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 12th 2009, 04:33 PM
  4. Write the expressions in terms of cosine?
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 11th 2009, 11:27 AM
  5. Express in terms of cosine.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 16th 2008, 09:38 AM

/mathhelpforum @mathhelpforum