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Coordinates of the electric field vectors of a dipole

  • Thread starter JulienB
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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 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.
 

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Answers and Replies

  • #2
stevendaryl
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I think you're on the right track. However, it might be helpful to convince yourself of a few basic facts about gradients:

  • [itex]\nabla (\vec{p} \cdot \vec{r}) = \vec{p}[/itex] (if [itex]\vec{p}[/itex] is a constant vector)
  • [itex]\nabla f(r) = \frac{df}{dr} \hat{r}[/itex] (where [itex]\hat{r}[/itex] is a unit vector in the direction [itex]\vec{r}[/itex])
  • [itex]\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})[/itex]
 
  • #3
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@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.
 
  • #4
408
12
Now it looks like that:

[tex]
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}
[/tex]

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.
 
  • #5
SammyS
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Now it looks like that:
[tex]
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}
[/tex]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.
 
  • #6
408
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@SammyS hi Sammy and thanks for your answer. Then is that correct:

[tex]
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}
[/tex]

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


Julien.
 
  • #7
SammyS
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@SammyS hi Sammy and thanks for your answer. Then is that correct:

[tex]
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}
[/tex]

? 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.
 
  • #8
408
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@SammyS Thanks a lot! Then, if I understood correctly:

[tex]
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} \\
[/tex]

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

Thanks a lot for your precious help :)


Julien.
 
  • #9
SammyS
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Both ##\ p_y \text{ and } p_z \ ## are zero.
 
  • #10
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@SammyS Perfect, thanks a lot for all your help.


Julien.
 
  • #11
ehild
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##(\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##
$$\vec E = -\frac{1}{4 \pi \varepsilon_0} \frac{\vec p r^2 -3 (\vec p \cdot \vec r ) \vec r}{r^5}$$
 
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  • #12
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@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:

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

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


Julien.
 
  • #13
SammyS
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##(\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 \ .##
 
  • #14
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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.
 
  • #15
ehild
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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.
 
  • #16
SammyS
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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.
 
  • #17
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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.
 

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