- #1
JulienB
- 408
- 12
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 x-direction.
Calculate from the general relation E = -∇φ the coordinates Ex, Ey, Ez 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 "pj" 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 = rx + ry + rz 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 py = 0, but I think I cannot say what pz is equal too since the problem only says that p goes in the direction of the x-axis.
[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 Ey = 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.