# I What is the continuous electric dipole distribution?

#### Mike400

An electric dipole is a system of two opposite point charges when their separation goes to zero and their charge goes to infinity in a way that the product of the charge and the separation remains finite.

Now how can we have a continuous electric dipole volume distribution from such a collection of point charges?

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#### PeroK

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An electric dipole is a system of two opposite point charges when their separation goes to zero and their charge goes to infinity in a way that the product of the charge and the separation remains finite.

Now how can we have a continuous electric dipole volume distribution from such a collection of point charges?
You can't physically, but you can have a physical scenario that can be modelled as a continuous distribution of perfect dipoles. The model is an approximation to a vast number of tiny physical dipoles.

In fact, you can't physically have a continuous distribution of charge: ultimately it's always a finite number of point charges.

And, also, no object can be a continuous distribution of mass: it's always a collection of elementary particles arranged in mostly empty space.

#### Mike400

You can't physically, but you can have a physical scenario that can be modelled as a continuous distribution of perfect dipoles. The model is an approximation to a vast number of tiny physical dipoles.

In fact, you can't physically have a continuous distribution of charge: ultimately it's always a finite number of point charges.

And, also, no object can be a continuous distribution of mass: it's always a collection of elementary particles arranged in mostly empty space.
I am also just talking mathematically. What I am trying to say is, mathematically if you want a continuous volume charge distribution, it must be composed of three dimensional volume elements, and not point charges. Similarly in order to have a continuous electric dipole volume distribution, it must be composed of three dimensional volume elements. But dipoles by definition are one dimensional (two point charges separated by a distance).

So how can we have a continuous electric dipole volume distribution?

#### PeroK

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I am also just talking mathematically. What I am trying to say is, mathematically if you want a continuous volume charge distribution, it must be composed of three dimensional volume elements, and not point charges. Similarly in order to have a continuous electric dipole volume distribution, it must be composed of three dimensional volume elements. But dipoles by definition are one dimensional (two point charges separated by a distance).

So how can we have a continuous electric dipole volume distribution?
By defintion a perfect dipole is a 3D vector at a point is space. Any physical dipole is only a approximation to this. But, if the dipoles are tiny and vast in number, then the two models are close enough.

A continous dipole distribution is, therefore, a vector field; whereas, a continuous charge distribution is a scalar field.

Much of physics, in terms of its use of calculus, boils down to this issue of a continuous approximation to a discrete, finite reality.

#### Mike400

By defintion a perfect dipole is a 3D vector at a point is space. Any physical dipole is only a approximation to this. But, if the dipoles are tiny and vast in number, then the two models are close enough.

A continous dipole distribution is, therefore, a vector field; whereas, a continuous charge distribution is a scalar field.

Much of physics, in terms of its use of calculus, boils down to this issue of a continuous approximation to a discrete, finite reality.
Then is the infinitesimal element of a continuous dipole distribution three dimensional?

#### PeroK

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Then is the infinitesimal element of a continuous dipole distribution three dimensional?
Yes. Just like $d\vec{E}$ for the electric field.

Which is really just the field at a point times the usual volume element at that point; which, itself, is just the limit of the average field on a volume times the finite volume.

#### Mike400

Yes. Just like $d\vec{E}$ for the electric field.

Which is really just the field at a point times the usual volume element at that point; which, itself, is just the limit of the average field on a volume times the finite volume.
The polarization density is:

$\mathbf{M}=\dfrac{d\mathbf{p}}{dV}$

Therefore in an element volume, $d\mathbf{p}=dq\ d\mathbf{l}$ vanishes to the third order.

That is, $dq$ vanishes to the second order and hence must be a quantity over some area of volume element of continuous dipole distribution. Am I correct?

#### PeroK

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The polarization density is:

$\mathbf{M}=\dfrac{d\mathbf{p}}{dV}$

Therefore in an element volume, $d\mathbf{p}=dq\ d\mathbf{l}$ vanishes to the third order.

That is, $dq$ vanishes to the second order and hence must be a quantity over some area of volume element of continuous dipole distribution. Am I correct?
I don't really follow what you are doing. Once we model a polarised object as a continuous dipole distribution, the concept of a large number of physical dipoles has gone. Instead we have the potential associated with each perfect dipole: $$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^2}$$
And, when we integrate this over a polarised object, we find that the polarised object is equivalent (in terms of the electric field it creates) to a certain volume charge and a certain surface charge.

Now, we have a model for a polarised object that is a) clearly physically very different (from the reality) and yet b) electrostatically equivalent.

#### Mike400

I don't really follow what you are doing. Once we model a polarised object as a continuous dipole distribution, the concept of a large number of physical dipoles has gone. Instead we have the potential associated with each perfect dipole: $$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^2}$$
And, when we integrate this over a polarised object, we find that the polarised object is equivalent (in terms of the electric field it creates) to a certain volume charge and a certain surface charge.

Now, we have a model for a polarised object that is a) clearly physically very different (from the reality) and yet b) electrostatically equivalent.
What do you mean by electrostatically equivalent?

#### PeroK

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What do you mean by electrostatically equivalent?
Creates the same electric field.

#### Mike400

I don't really follow what you are doing. Once we model a polarised object as a continuous dipole distribution, the concept of a large number of physical dipoles has gone. Instead we have the potential associated with each perfect dipole: $$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^2}$$
And, when we integrate this over a polarised object, we find that the polarised object is equivalent (in terms of the electric field it creates) to a certain volume charge and a certain surface charge.

Now, we have a model for a polarised object that is a) clearly physically very different (from the reality) and yet b) electrostatically equivalent.
We can view the charge distribution as an infinite number of element volume charges. So when modelling a polarised object as a continuous dipole distribution, what is the reason which prevents us from viewing it as an infinite number of element volume dipoles?

#### PeroK

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We can view the charge distribution as an infinite number of element volume charges. So when modelling a polarised object as a continuous dipole distribution, what is the reason which prevents us from viewing it as an infinite number of element volume dipoles?
Mathematics can't handle infinities of that sort. The transition from a large finite number to a continous distribution is required. That's really what the integral calculus does.

If you had an infinite number of dipoles then the field would be infinite. Or, if all the dipoles had moment 0, then the field would be 0.

#### PeroK

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We can view the charge distribution as an infinite number of element volume charges. So when modelling a polarised object as a continuous dipole distribution, what is the reason which prevents us from viewing it as an infinite number of element volume dipoles?
Fundamentally, for some reason that I cannot understand, you don't seem to be able to accept the concept of a perfect dipole as an approximation/model for a tiny physical dipole.

That's what this whole thing boils down to. A perfect dipole is a decidedly unphysical concept; but, so is a point charge or a point mass.

#### Mike400

The polarization density is:

$\mathbf{M}=\dfrac{d\mathbf{p}}{dV}$

Therefore in an element volume, $d\mathbf{p}=dq\ d\mathbf{l}$ vanishes to the third order.

That is, $dq$ vanishes to the second order and hence must be a quantity over some area of volume element of continuous dipole distribution. Am I correct?
Here I explain what I mean by the above post.

Let a small element volume dipole be having volume $\Delta{V}=\Delta{x}\ \Delta{y}\ \Delta{l}$

Polarization density at a point inside it will be:

$\mathbf{M}=\dfrac{\Delta{\mathbf{p}}}{\Delta{V}}=\dfrac{q\ \Delta{l}}{\Delta{x}\ \Delta{y}\ \Delta{l}}$

$|\mathbf{M}|=\dfrac{q}{\Delta{x}\ \Delta{y}}$

In the limit as $\Delta{x} \to 0$, $\Delta{y} \to 0$, $\Delta{l} \to 0$

$\Delta{x}, \Delta{y}, \Delta{l}$ vanishes to the first order
i.e. $\Delta{x}\ \Delta{y}$ vanishes to the second order
i.e. $q$ vanishes to the second order

Therefore it seems we can consider $q$ as a surface charge and $\lim \limits_{\Delta{x} \to 0} \lim \limits_{\Delta{y} \to 0} \dfrac{q}{\Delta{x}\ \Delta{y}} = |\mathbf{M}|$ as surface charge density.

Is this interpretation correct?

#### PeroK

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Here I explain what I mean by the above post.

Let a small element volume dipole be having volume $\Delta{V}=\Delta{x}\ \Delta{y}\ \Delta{l}$

Polarization density at a point inside it will be:

$\mathbf{M}=\dfrac{\Delta{\mathbf{p}}}{\Delta{V}}=\dfrac{q\ \Delta{l}}{\Delta{x}\ \Delta{y}\ \Delta{l}}$

$|\mathbf{M}|=\dfrac{q}{\Delta{x}\ \Delta{y}}$

In the limit as $\Delta{x} \to 0$, $\Delta{y} \to 0$, $\Delta{l} \to 0$

$\Delta{x}, \Delta{y}, \Delta{l}$ vanishes to the first order
i.e. $\Delta{x}\ \Delta{y}$ vanishes to the second order
i.e. $q$ vanishes to the second order

Therefore it seems we can consider $q$ as a surface charge and $\lim \limits_{\Delta{x} \to 0} \lim \limits_{\Delta{y} \to 0} \dfrac{q}{\Delta{x}\ \Delta{y}} = |\mathbf{M}|$ as surface charge density.

Is this interpretation correct?
No. I don't really follow what you are doing, but the wrong answer comes out in any case.

If you have $\vec{M}$ as the dipole moment per unit volume, then you should get a volume bound charge density of

$\rho_b = - \nabla \cdot \vec{M}$

And, as surface bound charge density of:

$\sigma_b = \vec{M} \cdot \hat{n}$

Where $\hat{n}$ is the normal to the surface.

Note that in general $\vec{M}$ is a function of position within the polarised object.

For example, for a uniformly polarised sphere the bound volume charge is zero and the bound surface charge is: $M\cos \theta$.

#### Mike400

I don't really follow what you are doing. Once we model a polarised object as a continuous dipole distribution, the concept of a large number of physical dipoles has gone. Instead we have the potential associated with each perfect dipole: $$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^2}$$
And, when we integrate this over a polarised object, we find that the polarised object is equivalent (in terms of the electric field it creates) to a certain volume charge and a certain surface charge.

Now, we have a model for a polarised object that is a) clearly physically very different (from the reality) and yet b) electrostatically equivalent.
Shouldn't it be $V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{dp} \cdot \vec{r}}{r^2}$ so that $V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{M} \cdot \vec{r}}{r^2} dV$ and by integrating we have $\displaystyle \dfrac{1}{4\pi \epsilon_0} \int_V \dfrac{\vec{M} \cdot \vec{r}}{r^2} dV$

#### PeroK

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Shouldn't it be $V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{dp} \cdot \vec{r}}{r^2}$ so that $V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{M} \cdot \vec{r}}{r^2} dV$ and by integrating we have $\displaystyle \dfrac{1}{4\pi \epsilon_0} \int_V \dfrac{\vec{M} \cdot \vec{r}}{r^2} dV$
Instead we have the potential associated with each perfect dipole: $$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^2}$$
To be precise, this is the potential of a perfect dipole at the origin; or, at least where $\vec{r}$ is the displacement from the dipole.

The potential for a volume of continuous dipole distribution would be:

$$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{\vec{M(\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2} dV'$$

#### Mike400

To be precise, this is the potential of a perfect dipole at the origin; or, at least where $\vec{r}$ is the displacement from the dipole.

The potential for a volume of continuous dipole distribution would be:

$$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \int \dfrac{\vec{M(\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2} dV'$$
How shall we derive potential for a volume of continuous dipole distribution $$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \int \dfrac{\vec{M(\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2} dV'$$ from potential of a perfect dipole $$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{p (\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2}$$

#### PeroK

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How shall we derive potential for a volume of continuous dipole distribution $$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \int \dfrac{\vec{M(\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2} dV'$$ from potential of a perfect dipole $$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{p (\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2}$$
You just integrate!

$M dV' \equiv p$

#### Mike400

You just integrate!

$M dV' \equiv p$
Doesn't it mean we are adding the tiny potentials due to each approximately perfect tiny dipoles in the limit as the number of tiny dipoles tends to infinity. But you said that the concept of a large number of physical dipoles is gone once we model a polarised object as a continuous dipole distribution.

"What is the continuous electric dipole distribution?"

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