Gauss Law and Electric Field of a Dipole

[Dr_Pill]

Hi,

I don't understand this:
You got a dipole, and a resulting electric field from it like this(hyperphysics):

As z approaches infinity, the field becomes zero.

But if you draw a gaussian surface round the charges, then the net enclosed charge is zero.So the electric field must be zero, no matter how I draw the surface.
So If i just draw a surface tiny enough to enclose both charges, my z is totally not going to infinity and the field is zero according to Gauss' LAw.

How is this compatible with each other?What is the pitfall in my thinking.

Thanks in advance.

 

[jtbell]

So the electric field must be zero, no matter how I draw the surface.

No, the integral of the electric field over the entire surface (the flux of the electric field) must be zero.

On some parts of the surface, the field points outwards, which contributes a positive amount to the flux. On other parts of the surface, the field points inwards, which contributes a negative amount to the flux. The positive and negative contributions add up to zero over the entire surface.

 

[vanhees71]

Of course, you have to use the full electric field due to your dipole and an electric field is a vector. By construction any partial wave corresponds to a charge distribution with total net charge 0 except the monopole term.

The dipol field is given by the potential
$$\Phi(\vec{x})=\frac{\vec{p} \cdot \vec{x}}{4 \pi r^3} \quad \text{with} \quad r=|\vec{x}|.$$
The field is thus given by
$$\vec{E}=-\vec{\nabla} \Phi=\frac{1}{4 \pi r^5} [3(\vec{p} \cdot \vec{x}) \vec{x}-r^2 \vec{p}].$$
Here, the dipol sits at the origin of the coordinate system. The most simple surface to use is a sphere of radius $a$ around the origin. We use spherical coordinates with the polar axis in direction of the electric moment, $\vec{p}$. Then we have
$$\mathrm{d}^2 \vec{F}=\mathrm{d} \vartheta \mathrm{d} \varphi a^2 \sin \vartheta \vec{e}_r$$
and thus
$$\int_{K_a} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \; \frac{p \sin \vartheta \cos \vartheta}{2 \pi a} =0.$$

 

[stevendaryl]

Hi,

I don't understand this:
You got a dipole, and a resulting electric field from it like this(hyperphysics):

As z approaches infinity, the field becomes zero.

But if you draw a gaussian surface round the charges, then the net enclosed charge is zero.So the electric field must be zero, no matter how I draw the surface.
So If i just draw a surface tiny enough to enclose both charges, my z is totally not going to infinity and the field is zero according to Gauss' LAw.

How is this compatible with each other?What is the pitfall in my thinking.

Thanks in advance.

As others have pointed out, Gauss' Law tells us that if a surface encloses total charge 0, then the AVERAGE of the field over the surface is zero. That doesn't mean that it is zero at every point. In the case of a dipole, the part of surface nearest to the + charge will have a positive flux through it, and the part of the surface nearest to the - charge will have a negative flux through it.

Gauss' Law doesn't really help much in figuring out what the electric field is for a distribution of charges unless there is some kind of symmetry involved. So for instance, a flat uniform sheet of charge, a line of charge, a spherical shell of charge.

 

[sardar]

Hello,

Thank you for bringing up this question about Gauss Law and the electric field of a dipole. It is important to understand that Gauss Law is a mathematical relationship that describes the behavior of electric fields. In the case of a dipole, the electric field is indeed zero at infinity, as shown in the diagram you provided.

However, when we apply Gauss Law to this situation, we are considering a closed surface around the dipole. This surface can be any size, including a tiny surface that encloses both charges. In this case, the net enclosed charge is indeed zero, but this does not mean that the electric field is zero everywhere on the surface.

In fact, Gauss Law tells us that the electric field is still present on the surface, but it is balanced out by the opposite charges of the dipole. This is why the net enclosed charge is zero, but the electric field is not. This is not a pitfall in your thinking, but rather a fundamental principle of Gauss Law and the behavior of electric fields.

I hope this explanation helps clarify the relationship between Gauss Law and the electric field of a dipole. If you have any further questions, please don't hesitate to ask. As scientists, it is important for us to continue questioning and seeking understanding in order to further our knowledge and discoveries. Thank you for your curiosity and interest in this topic.