Direction of dipole moment in e-field of another dipole

  • Thread starter JulienB
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  • #1
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Homework Statement



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##?

Homework 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})##

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.
 

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Answers and Replies

  • #2
408
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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)...
 
  • #3
TSny
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A second electric dipole ##\vec{p}_2## is placed at location ##(x_0, 0, y_0)##.
##(x_0, 0, y_0)## ⇒ ##(x_0, 0, z_0)##?
##\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}}##

##\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})##
The z-component of ##E_1## is incomplete. Note that ##z_0## occurs in both the numerator and denominator of ##\phi_1##.
 
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  • #4
408
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Hi @TSny and thanks for your answer. Oh right... That's unfortunately one of my common mistakes -_- Thanks for pointing that one out!

Julien.
 

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