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Confusion about dipoles

  1. Nov 23, 2014 #1
    Hi everybody. I noticed that I don't understand dipoles correctly. I just calculated the field of a dipole. There are two charges, one positive the other negative, in the distance of d.
    I got for the electric field: [tex]E=\frac{q}{4 \pi \epsilon} \binom{d}{0}[/tex] The problem is, that there is only the x-component left. I don't see how the y-component got cancelled. I mean it makes sense from the mathematics but I can't visualize how this happens. Also pictures of dipoles you find in books or on the internet showing electric field lines that have more than just one component. See also here: http://en.wikipedia.org/wiki/Dipole

    What did I understand wrong?

    kind regards.
  2. jcsd
  3. Nov 23, 2014 #2

    Simon Bridge

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    Pick a point on the y axis - anywhere - and work out the electric field there. Draw an arrow in there for the field.
    Pick another point, not on the y axis, do it again. After a while you will have built up a picture of the electric field.

    As to your equation, I have: $$\vec E = \frac{1}{4\pi\epsilon_0}\frac{1}{r^3}\big[3(\vec p \cdot \vec r) \hat r - \vec p\big]$$ re http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf ... eq13.
    For your case: ##\vec p = q(d,0,0)^t## I suspect you have ##\vec r = (x,y,0)^t## ...

    Notice that it is 3D, and the electric field strength decreases with distance from the dipole - but yours is 2D and the strength does not change with distance?
  4. Nov 24, 2014 #3
    Thanks for your reply. I see now what my problem was. I had only one component because I just looked for the field on the y-axis that its exactly on x=d/2. So in the exact mid of the two charges with respect to the x-axis. So I just didn't calculated the complete field for every point on the plane. Sometimes I don't see my most obvious mistakes. Thank you.
    edit: And yeah, the given field doesn't change with distance, I just forgot to append the '1/z^3' to the formula and now I can't edit it?

    Kind regards.
    Last edited: Nov 24, 2014
  5. Nov 24, 2014 #4
    Here's pic of E-field lines you're interested:
    One can clearly see where horizontal component of E-field dissapears.
  6. Nov 24, 2014 #5

    Simon Bridge

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    The dipole would normally be centered on the origin so how is x=d/2 the exact mid? I dont see how you get a z with only a 2D displaceme t vector.
    How about rewriting the equation properly and stati g clearly what it is for?
    That should help you get clear in you mind.
  7. Nov 24, 2014 #6


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    Let's derive the static-dipole field from scratch. The idea is to look at two equal opposite charges at a point which is far from the distance of the charges. Let's orient the dipole along the [itex]z[/itex] axis of a Cartesian coordinate system. The positive charge sits at [itex](0,0,-d/2)[/itex] and the negative charge ate [itex](0,0,+d/2)[/itex], where [itex]d[/itex] is the distance between the charges. The dipole moment then is [itex]\vec{p}=q d \vec{e}_z[/itex]. I work in Heaviside-Lorentz units.

    The potential of the two charges is
    [tex]\Phi(\vec{r})=\frac{q}{4 \pi} \left (\frac{1}{\sqrt{x^2+y^2+(z+d/2)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-d/2)^2}} \right).[/tex]
    Now for [itex]r=\sqrt{x^2+y^2+z^2} \gg d[/itex] we can expand this to first-order in [itex]d[/itex]. This gives
    [tex]\Phi(\vec{r}) \simeq \frac{1}{4 \pi} \vec{p} \cdot \vec{\nabla} \frac{1}{r}=\frac{\vec{p} \cdot \vec{r}}{4 \pi r^3}.[/tex]
  8. Nov 24, 2014 #7
    Sry, I thought of one charge on the origin and the other at +d. But it's better to put the charges on -d/2 and d/2 of course.
    And I used the x-z plane, so this might be confusing, sry for that.

    Thanks, this helped also a lot for my understanding. Taylor expansion is something I learned but I'm not sure when to use it.
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