# Coordinates of the electric field vectors of a dipole

## 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:

$$\phi (\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{|\vec{r}|^3}$$

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

$$\vec{E} = -\nabla \phi (\vec{r})$$

## 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:

$$\vec{E} = -\nabla \phi (\vec{r}) = -(\frac{\partial \phi (\vec{r})}{\partial r_x} \vec{e_x} + \frac{\partial \phi (\vec{r})}{\partial r_y} \vec{e_y} + \frac{\partial \phi (\vec{r})}{\partial r_z} \vec{e_z}) \\ \implies E_j = -\frac{\partial \phi (\vec{r})}{\partial r_j} \vec{e_j}$$

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:
$$E_j = \frac{-1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{\vec{p} \cdot \vec{r}}{r^3}) \vec{e_j}$$

Then I do the dot product and develop the magnitude underneath:

$$E_j = \frac{-1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{p_x \cdot r_x + p_y \cdot r_y + p_z \cdot r_z}{(r_x + r_y + r_z)^{\frac{3}{2}}}) \vec{e_j}$$

I can differentiate the top and I can take "pj" out:

$$E_j = \frac{-p_j}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{1}{(r_x + r_y + r_z)^{\frac{3}{2}}}) \vec{e_j}$$

I substitute with u = rx + ry + rz and differentiate:

$$E_j = \frac{-p_j}{4 \pi \varepsilon_0} \frac{\partial}{\partial u} \frac{1}{u^{\frac{3}{2}}} \frac{\partial u}{\partial r_j} \vec{e_j} \\ = \frac{-p_j}{4 \pi \varepsilon_0} \frac{-3}{2u^{\frac{5}{2}}} \frac{\partial u}{\partial r_j} \vec{e_j} \\ = \frac{p_j}{4 \pi \varepsilon_0} \frac{3}{2r^5} 2r_j \vec{e_j} \\ = \frac{p_j}{4 \pi \varepsilon_0} \frac{3}{r^5} r_j \vec{e_j}$$

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.

$$E_x = \frac{3 p_x r_x}{4 \pi \varepsilon_0 r^5} \vec{e_x} \\ E_y = 0 \\ E_z = \frac{3 p_z r_z}{4 \pi \varepsilon_0 r^5} \vec{e_z} \\$$

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?

Julien.

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stevendaryl
Staff Emeritus
I think you're on the right track. However, it might be helpful to convince yourself of a few basic facts about gradients:

• $\nabla (\vec{p} \cdot \vec{r}) = \vec{p}$ (if $\vec{p}$ is a constant vector)
• $\nabla f(r) = \frac{df}{dr} \hat{r}$ (where $\hat{r}$ is a unit vector in the direction $\vec{r}$)
• $\nabla \frac{\vec{p} \cdot \vec{r}}{r^3} = \frac{\nabla (\vec{p} \cdot \vec{r})}{r^3} +(\vec{p} \cdot \vec{r})(\nabla \frac{1}{r^3})$

@stevendaryl thanks a lot for your answer. It looks a lot like the product rule, can I think of it like that to memorize it? I'm soon gonna post a correct answer baser on your indications.

Julien.

Now it looks like that:

$$E_j = - \frac{\partial}{\partial r_j} (\frac{1}{4 \pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}) \vec{e_j} = - \frac{1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} \frac{\vec{p} \cdot \vec{r}}{r^3} \vec{e_j} \\ = - \frac{1}{4 \pi \varepsilon_0} (\frac{\vec{p}}{r^3} + \vec{p} \cdot \vec{r} \frac{\partial}{\partial r_j} \frac{1}{r^3}) \vec{e_j} \\ = - \frac{\vec{p}}{4 \pi \varepsilon_0} (\frac{1}{r^3} + \frac{3 r_j}{r^5} \vec{r}) \vec{e_j} \\ = - \frac{\vec{p}}{4 \pi \varepsilon_0 r^3} (1 + \frac{3 r_j}{r^2} \vec{r}) \vec{e_j}$$

Is that any better? It looks a bit crazy :)

Then I could replace all rj with an expression involving r and θ, that at least would make sense!

Julien.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Now it looks like that:
$$E_j = - \frac{\partial}{\partial r_j} (\frac{1}{4 \pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}) \vec{e_j} = - \frac{1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} \frac{\vec{p} \cdot \vec{r}}{r^3} \vec{e_j} \\ = - \frac{1}{4 \pi \varepsilon_0} (\frac{\vec{p}}{r^3} + \vec{p} \cdot \vec{r} \frac{\partial}{\partial r_j} \frac{1}{r^3}) \vec{e_j} \\ = - \frac{\vec{p}}{4 \pi \varepsilon_0} (\frac{1}{r^3} + \frac{3 r_j}{r^5} \vec{r}) \vec{e_j} \\ = - \frac{\vec{p}}{4 \pi \varepsilon_0 r^3} (1 + \frac{3 r_j}{r^2} \vec{r}) \vec{e_j}$$Is that any better? It looks a bit crazy :)

Then I could replace all rj with an expression involving r and θ, that at least would make sense!

Julien.
There is a problem.

You have a term with the product $\ \vec{p}\, \vec{e}_j \$.

As steven said, $\ \nabla (\vec{p} \cdot \vec{r}) = \vec{p} \,,\$ so the j component of this gradient is simply $\ p_j\$.

Also, you mishandled $\ \vec{p}\cdot\vec r\$. It disappeared after the second line.

@SammyS hi Sammy and thanks for your answer. Then is that correct:

$$E_j = - \frac{1}{4 \pi \varepsilon_0 r^3} (p_j + \vec{p} \cdot \vec{r} \frac{3 r_j}{r^2}) \vec{e_j}$$

? If yes, can it be simplified? This is all pretty confusing and ugly looking.

Julien.

SammyS
Staff Emeritus
Homework Helper
Gold Member
@SammyS hi Sammy and thanks for your answer. Then is that correct:

$$E_j = - \frac{1}{4 \pi \varepsilon_0 r^3} (p_j + \vec{p} \cdot \vec{r} \frac{3 r_j}{r^2}) \vec{e_j}$$

? If yes, can it be simplified? This is all pretty confusing and ugly looking.

Julien.
Looks good.

That's for $\ \vec p \$ in an arbitrary direction. The statement of the problem has $\ \vec p \$ in the x-direction.

• JulienB
@SammyS Thanks a lot! Then, if I understood correctly:

$$E_j = - \frac{1}{4 \pi \varepsilon_0 r^3} ( p_j + p_x \cdot r_x \frac{3 r_j}{r^2}) \vec{e_j} \\ \implies E_x = - \frac{p_x}{4 \pi \varepsilon_0 r^3} (1 + \frac{3 r_x^2}{r^2}) \vec{e_x} \\ \mbox{and } E_y = - \frac{1}{4 \pi \varepsilon_0 r^3} (p_x \cdot r_x \frac{3 r_y}{r^2}) \vec{e_y} \\ \mbox{and } E_z = - \frac{1}{4 \pi \varepsilon_0 r^3} (p_z + p_x \cdot r_x \frac{3 r_z}{r^2}) \vec{e_z} \\$$

Is that correct? I am still unsure whether pz = 0 or not. What do you think?

Thanks a lot for your precious help :)

Julien.

SammyS
Staff Emeritus
Homework Helper
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Both $\ p_y \text{ and } p_z \$ are zero.

• JulienB
@SammyS Perfect, thanks a lot for all your help.

Julien.

ehild
Homework Helper
$(\vec p\cdot \vec r)$ is a scalar, it dos not have x, y, z components.
Accordingly, the electric field vector is $$\vec E = - \frac{1}{4 \pi \varepsilon_0 r^3} (\vec p +( \vec{p} \cdot \vec{r}) \frac{3 \vec r}{r^2})$$.

If a function depends only on $\vec r$ you get the gradient easier if you formally differentiate with respect to $\vec r$.
The potential is $$\phi (\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{|\vec{r}|^3}$$
$|\vec r |=\sqrt{(\vec r)^2}$. Take the negative derivative of $$\frac{1}{4 \pi \varepsilon_0}\frac{(\vec{p} \cdot \vec{r})}{((\vec{r})^2)^{3/2}}$$
with respect to $\vec r$, It becomes
$$-\frac{1}{4 \pi \varepsilon_0} \frac{\vec p ( (\vec r)^2)^{3/2} -3 (\vec p \cdot \vec r )((\vec r)^2)^{1/2} \vec r}{((\vec r)^2)^3}$$

Rewriting $(\vec r)^2=r^2$
$$\vec E = -\frac{1}{4 \pi \varepsilon_0} \frac{\vec p r^2 -3 (\vec p \cdot \vec r ) \vec r}{r^5}$$

Last edited:
• BvU
@ehild Hi ehild and thanks for your participation to the topic :)

I'm not sure what you mean with the x,y,z components of p⋅r. I also wrote it as a scalar just doing the dot product, and it turns out like that:

$$\vec{p} \cdot \vec{r} = p_x r_x + p_y r_y + p_z r_z = p_x r_x \mbox{ since p_y = p_z = 0}$$

I'll take a look at the rest as soon as possible (I'm at work ). Are you saying my previous result was wrong, or are you proposing an alternative way to solve it, possibly more efficiently?

Julien.

• SammyS
SammyS
Staff Emeritus
Homework Helper
Gold Member
$(\vec p\cdot \vec r)$ is a scalar, it dos not have x, y, z components.
Accordingly, the electric field vector is $$\vec E = - \frac{1}{4 \pi \varepsilon_0 r^3} (\vec p +( \vec{p} \cdot \vec{r}) \frac{3 \vec r}{r^2})$$.
...

Rewriting $(\vec r)^2=r^2$
$$\vec E = -\frac{1}{4 \pi \varepsilon_0} \frac{\vec p r^2 -3 (\vec p \cdot \vec r ) \vec r}{r^5}$$
This is equivalent to Julien's for the case $\ \vec p = p_x \, \hat e_x \ .$

Julien did use $\ "\cdot "\$ for simple multiplication of $\ p_x \text{ times } r_x \ .$

Julien did use $\ "\cdot "\$ for simple multiplication of $\ p_x \text{ times } r_x \ .$
Should I not? Sometimes I am confused about the conventions about products of different sorts, it seems a bit like everyone kind of does what they want as long as it is understandable.

Julien.

ehild
Homework Helper
Are you saying my previous result was wrong, or are you proposing an alternative way to solve it, possibly more efficiently?

Julien.
Your end result was correct, I only wanted to show an alternative method to get the gradient of a function.
There was a mistake in your first post
$$E_j = \frac{-1}{4 \pi \varepsilon_0} \frac{\partial}{\partial r_j} (\frac{p_x \cdot r_x + p_y \cdot r_y + p_z \cdot r_z}{(r_x + r_y + r_z)^{\frac{3}{2}}}) \vec{e_j}$$
There should be squares of the coordinates in the denominator. I don't see why you write the coordinates x, y, z as rx, ry, rz. As for the dot product $(\vec p \cdot \vec r )$ I might have misread something.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Should I not? Sometimes I am confused about the conventions about products of different sorts, it seems a bit like everyone kind of does what they want as long as it is understandable.

Julien.
It was proper.

There should be squares of the coordinates in the denominator. I don't see why you write the coordinates x, y, z as rx, ry, rz.
Yes I forgot the squares indeed. As for $r_j$, I just do that to make sure I don't get confused about which coordinates I am talking about. I guess it is pretty obvious though.

Thanks a lot to everybody for your contribution! ;) You guys are always so helpful!

Julien.