- #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 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 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 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.