What is the continuous electric dipole distribution?

In summary: 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:...
  • #1
Mike400
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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?
 
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  • #2
Mike400 said:
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 modeled 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.
 
  • #3
PeroK said:
You can't physically, but you can have a physical scenario that can be modeled 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?
 
  • #4
Mike400 said:
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 continuous 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.
 
  • #5
PeroK said:
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 continuous 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?
 
  • #6
Mike400 said:
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.
 
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  • #7
PeroK said:
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?
 
  • #8
Mike400 said:
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.
 
  • #9
PeroK said:
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?
 
  • #10
Mike400 said:
What do you mean by electrostatically equivalent?

Creates the same electric field.
 
  • #11
PeroK said:
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?
 
  • #12
Mike400 said:
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 continuous 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.
 
  • #13
Mike400 said:
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.
 
  • #14
Mike400 said:
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?
 
  • #15
Mike400 said:
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##.
 
  • #16
PeroK said:
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##
 
  • #17
Mike400 said:
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 said:
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:

$$V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{\vec{M(\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2} dV'$$
 
  • #18
PeroK said:
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 $$V(\vec{r}) = \dfrac{1}{4\pi \epsilon_0} \dfrac{\vec{p (\vec{r'})} \cdot (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^2}$$
 
  • #19
Mike400 said:
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##
 
  • #20
PeroK said:
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.
 

1. What is a continuous electric dipole distribution?

A continuous electric dipole distribution refers to a distribution of electric dipoles that are continuously present throughout a given region of space. This means that there is no gap or break in the distribution, and the dipoles are evenly distributed throughout the region.

2. How is a continuous electric dipole distribution different from a discrete electric dipole distribution?

A continuous electric dipole distribution differs from a discrete electric dipole distribution in that the dipoles in a continuous distribution are spread out and evenly distributed, while in a discrete distribution, the dipoles are located at specific points or positions.

3. What are the applications of a continuous electric dipole distribution?

A continuous electric dipole distribution has various applications in physics and engineering, including in the study of electromagnetic fields, the behavior of dielectric materials, and the design of antennas and other electronic devices.

4. How is a continuous electric dipole distribution represented mathematically?

A continuous electric dipole distribution can be represented mathematically using the dipole moment density, which is a vector quantity that describes the strength and direction of the dipoles at any point in the distribution. This can be expressed as a function of position in three-dimensional space.

5. What factors affect the strength of a continuous electric dipole distribution?

The strength of a continuous electric dipole distribution is affected by various factors, including the distance between the dipoles, the magnitude of the individual dipole moments, and the orientation of the dipoles relative to each other. Additionally, the presence of external electric fields can also influence the strength of the distribution.

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