Why is the y-component of the electric field in a dipole cancelled out?

[Raman student]

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: $$E=\frac{q}{4 \pi \epsilon} \binom{d}{0}$$ 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.

 

[Simon Bridge]

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?

 

[Raman student]

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.

 

[zoki85]

Here's pic of E-field lines you're interested:

One can clearly see where horizontal component of E-field dissapears.

 

[Simon Bridge]

The dipole would normally be centered on the origin so how is x=d/2 the exact mid? I don't 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.

 

[vanhees71]

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 $z$ axis of a Cartesian coordinate system. The positive charge sits at $(0,0,-d/2)$ and the negative charge ate $(0,0,+d/2)$, where $d$ is the distance between the charges. The dipole moment then is $\vec{p}=q d \vec{e}_z$. I work in Heaviside-Lorentz units.

The potential of the two charges is
$$\Phi(\vec{r})=\frac{q}{4 \pi} \left (\frac{1}{\sqrt{x^2+y^2+(z+d/2)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-d/2)^2}} \right).$$
Now for $r=\sqrt{x^2+y^2+z^2} \gg d$ we can expand this to first-order in $d$. This gives
$$\Phi(\vec{r}) \simeq \frac{1}{4 \pi} \vec{p} \cdot \vec{\nabla} \frac{1}{r}=\frac{\vec{p} \cdot \vec{r}}{4 \pi r^3}.$$

 

[Raman student]

The dipole would normally be centered on the origin so how is x=d/2 the exact mid? I don't 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.

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.

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 $z$ axis of a Cartesian coordinate system. The positive charge sits at $(0,0,-d/2)$ and the negative charge ate $(0,0,+d/2)$, where $d$ is the distance between the charges. The dipole moment then is $\vec{p}=q d \vec{e}_z$. I work in Heaviside-Lorentz units.

The potential of the two charges is
$$\Phi(\vec{r})=\frac{q}{4 \pi} \left (\frac{1}{\sqrt{x^2+y^2+(z+d/2)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-d/2)^2}} \right).$$
Now for $r=\sqrt{x^2+y^2+z^2} \gg d$ we can expand this to first-order in $d$. This gives
$$\Phi(\vec{r}) \simeq \frac{1}{4 \pi} \vec{p} \cdot \vec{\nabla} \frac{1}{r}=\frac{\vec{p} \cdot \vec{r}}{4 \pi r^3}.$$

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