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Coordinates of the electric field vectors of a dipole

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    [tex]
    \vec{E} = -\nabla \phi (\vec{r})
    [/tex]

    3. 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.
     

    Attached Files:

  2. jcsd
  3. May 5, 2016 #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]
     
  4. May 5, 2016 #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.
     
  5. May 5, 2016 #4
    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.
     
  6. May 5, 2016 #5

    SammyS

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    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.
     
  7. May 5, 2016 #6
    @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.
     
  8. May 5, 2016 #7

    SammyS

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    Looks good.

    That's for ##\ \vec p \ ## in an arbitrary direction. The statement of the problem has ##\ \vec p \ ## in the x-direction.
     
  9. May 5, 2016 #8
    @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.
     
  10. May 5, 2016 #9

    SammyS

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    Both ##\ p_y \text{ and } p_z \ ## are zero.
     
  11. May 5, 2016 #10
    @SammyS Perfect, thanks a lot for all your help.


    Julien.
     
  12. May 5, 2016 #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##
    $$\vec E = -\frac{1}{4 \pi \varepsilon_0} \frac{\vec p r^2 -3 (\vec p \cdot \vec r ) \vec r}{r^5}$$
     
    Last edited: May 5, 2016
  13. May 6, 2016 #12
    @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.
     
  14. May 6, 2016 #13

    SammyS

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    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 \ .##
     
  15. May 6, 2016 #14
    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.
     
  16. May 6, 2016 #15

    ehild

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    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.
     
  17. May 6, 2016 #16

    SammyS

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    It was proper.
     
  18. May 6, 2016 #17
    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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