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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 zeropoint and shows in the zdirection. 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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