1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Direction of dipole moment in e-field of another dipole

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm trying to solve a problem about dipoles, but there is something I don't quite get about it. Well, first here is the problem:

    An electric dipole ##\vec{p}_1## is placed at the zero-point and shows in the z-direction. A second electric dipole ##\vec{p}_2## is placed at location ##(x_0, 0, y_0)##. In which direction does the dipole ##\vec{p}_2## show in the electric field of ##\vec{p}_1##?

    2. Relevant equations

    Dipole moment: ##\vec{p} = q \cdot \vec{d}##
    Potential of a dipole: ##\phi (\vec{r}) = k \cdot \frac{\vec{r} \cdot \vec{p}}{r^3}##
    Electric potential energy of a dipole: ##W_{pot} = - \vec{p}_2 \cdot \vec{E}##
    Electric field: ##\vec{E} (\vec{r}) = - \nabla \phi (\vec{r})##

    3. The attempt at a solution

    People who've read me before on this forum know that I loooove to draw a picture of the problems, so I've attached one as usual to this post. To solve the problem, my thinking was the following:

    I know that the potential energy of ##\vec{p}_2## will be minimal when it reaches its most stable position with respect to the electric field ##\vec{E}_1## of ##\vec{p}_1## at position ##\vec{r}##. That is, ##W_{pot} = - \vec{p}_2 \cdot \vec{E}_1 (\vec{r})## will be minimal when ##\vec{p}_2## is parallel to ##\vec{E}_1(\vec{r})##.

    In order to find what ##\vec{E}_1 (\vec{r})## is, I first calculate the potential of ##\vec{p}_1## at position ##\vec{r}## and get:

    ##\phi_1 (\vec{r}) = k \cdot \frac{\vec{r} \cdot \vec{p}}{r^3} = k \cdot \frac{z_0 \cdot p_z}{(x_0^2 + z_0^2)^{3/2}}##

    since ##p_x = p_y = y_0 = 0##. Then the electric field of ##\vec{p}_1## is simply:

    ##\vec{E}_1 (\vec{r}) = - \nabla \phi_1 (\vec{r}) = (3k \cdot x_0 \cdot z_0 \cdot p_z \cdot r^{-5}, 0, 3k \cdot z_0^2 \cdot p_z \cdot r^{-5})##
    ##= 3k \cdot z_0 \cdot p_z \cdot r^{-5} \cdot (x_0, 0, z_0)##
    ##= 3k \cdot z_0 \cdot p_z \cdot r^{-5} \cdot \vec{r}##

    And here comes a big surprise to me: ##\vec{E}_1 (\vec{r})## seems to be parallel to ##\vec{r}##! How is that possible? When I look at the picture I drew, that can't be true. Where is my mistake?

    I'm looking forward to reading you, thanks a lot in advance for your answers.


    Julien.
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2016 #2
    Maybe I have found my mistake. Could it be that I have to consider two superposing electric fields, one created by the positive charge and one created by the negative charge and add them up?

    Edit: Aah probably not since I use the definitions for ##\vec{p}## and not for ##\vec{d}## (distance between the two charges)...
     
  4. Sep 3, 2016 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    ##(x_0, 0, y_0)## ⇒ ##(x_0, 0, z_0)##?
    The z-component of ##E_1## is incomplete. Note that ##z_0## occurs in both the numerator and denominator of ##\phi_1##.
     
  5. Sep 3, 2016 #4
    Hi @TSny and thanks for your answer. Oh right... That's unfortunately one of my common mistakes -_- Thanks for pointing that one out!

    Julien.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Direction of dipole moment in e-field of another dipole
Loading...