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Homework Statement
Hi everybody! I might have solved that homework but I struggle to properly understand some steps, especially concerning the gradient and partial differentiation:
The potential Φ(r) of an electric dipole located at the origin of a coordinate system is given by:
[tex]
\phi (\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{\vec{r}^3}
[/tex]
where p is the electric dipole moment and r is the vector from the dipole to the point from which the potential is searched (see attached pic). The electric dipole moment is oriented in the xdirection.
Calculate from the general relation E = ∇φ the coordinates E_{x}, E_{y}, E_{z} of the electric field vector E.
Homework Equations
[tex]
\vec{E} = \nabla \phi (\vec{r})
[/tex]
The Attempt at a Solution
To begin with, I imagine that what they mean with "coordinates of the vector" is basically equivalent to "components", right? Here is how I started:
[tex]
\vec{E} = \nabla \phi (\vec{r}) = (\frac{\partial \phi (\vec{r})}{\partial r_x} \vec{e_x} + \frac{\partial \phi (\vec{r})}{\partial r_y} \vec{e_y} + \frac{\partial \phi (\vec{r})}{\partial r_z} \vec{e_z}) \\
\implies E_j = \frac{\partial \phi (\vec{r})}{\partial r_j} \vec{e_j}
[/tex]
Is that correct? My problem here is that I never really learned how to do a gradient (I guess it's time to learn!). So I've done the following steps with some help from the internet.
First we can remove the "constant terms" from the partial differential:
[tex]
E_j = \frac{1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{\vec{p} \cdot \vec{r}}{r^3}) \vec{e_j}
[/tex]
Then I do the dot product and develop the magnitude underneath:
[tex]
E_j = \frac{1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{p_x \cdot r_x + p_y \cdot r_y + p_z \cdot r_z}{(r_x + r_y + r_z)^{\frac{3}{2}}}) \vec{e_j}
[/tex]
I can differentiate the top and I can take "p_{j}" out:
[tex]
E_j = \frac{p_j}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{1}{(r_x + r_y + r_z)^{\frac{3}{2}}}) \vec{e_j}
[/tex]
I substitute with u = r_{x} + r_{y} + r_{z} and differentiate:
[tex]
E_j = \frac{p_j}{4 \pi \varepsilon_0} \frac{\partial}{\partial u} \frac{1}{u^{\frac{3}{2}}} \frac{\partial u}{\partial r_j} \vec{e_j} \\
= \frac{p_j}{4 \pi \varepsilon_0} \frac{3}{2u^{\frac{5}{2}}} \frac{\partial u}{\partial r_j} \vec{e_j} \\
= \frac{p_j}{4 \pi \varepsilon_0} \frac{3}{2r^5} 2r_j \vec{e_j} \\
= \frac{p_j}{4 \pi \varepsilon_0} \frac{3}{r^5} r_j \vec{e_j}
[/tex]
Then I know that p_{y} = 0, but I think I cannot say what p_{z} is equal too since the problem only says that p goes in the direction of the xaxis.
[tex]
E_x = \frac{3 p_x r_x}{4 \pi \varepsilon_0 r^5} \vec{e_x} \\
E_y = 0 \\
E_z = \frac{3 p_z r_z}{4 \pi \varepsilon_0 r^5} \vec{e_z} \\
[/tex]
I'm afraid this might be all wrong... :/ I didn't use the angle θ shown on the picture and that E_{y} = 0 doesn't really give me confidence. Maybe there is an easier way to solve that? What do you guys think?
Thanks a lot in advance for your answers.
Julien.
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