- #1

- 459

- 5

Why is

**D**called the "electric flux density vector"?
Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Swapnil
- Start date

- #1

- 459

- 5

Why is **D** called the "electric flux density vector"?

Last edited:

- #2

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,500

- 8

[tex]\vec{\nabla}\cdot\vec{D}=\rho[/tex]

- #3

- 5,820

- 1,123

- #4

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,500

- 8

Ah well, in that case back to Swapnil's question:

In addition to being the solution to Gauss' law, which I posted earlier, we can approach [itex]\vec{D}[/itex] a different way. Let's start with an electric field [itex]\vec{E}[/itex] in space in which there are free charges. It satisfies the equation [itex]\nabla\cdot\vec{E}=\rho/\epsilon_0[/itex] (which is just Gauss' law under these special circumstances). Now bring into the space a polarizable material. The electrons are pushed around to create another charge density, called the bound charge density [itex]\rho_B[/itex], which gives rise to a polarization density [itex]\vec{P}[/itex]. [itex]\vec{P}[/itex] is the solution to the differential equation [itex]\nabla\cdot\vec{P}=-\rho_B[/itex].

So now Gauss' law reads as follows.

[tex]\nabla\cdot\vec{E}=\frac{\rho + \rho_B}{\epsilon_0}[/tex]

[tex]\epsilon_0\nabla\cdot\vec{E}=\rho+\rho_B[/tex]

[tex]\nabla\cdot\epsilon_0\vec{E}=\rho-\nabla\cdot\vec{P}[/tex]

[tex]\nabla\cdot(\epsilon_0\vec{E}+\vec{P})=\rho[/tex]

Now**define** the operand of the divergence operator to be [itex]\vec{D}=\epsilon_0\vec{E}+\vec{P}[/itex], and you've got it.

In addition to being the solution to Gauss' law, which I posted earlier, we can approach [itex]\vec{D}[/itex] a different way. Let's start with an electric field [itex]\vec{E}[/itex] in space in which there are free charges. It satisfies the equation [itex]\nabla\cdot\vec{E}=\rho/\epsilon_0[/itex] (which is just Gauss' law under these special circumstances). Now bring into the space a polarizable material. The electrons are pushed around to create another charge density, called the bound charge density [itex]\rho_B[/itex], which gives rise to a polarization density [itex]\vec{P}[/itex]. [itex]\vec{P}[/itex] is the solution to the differential equation [itex]\nabla\cdot\vec{P}=-\rho_B[/itex].

So now Gauss' law reads as follows.

[tex]\nabla\cdot\vec{E}=\frac{\rho + \rho_B}{\epsilon_0}[/tex]

[tex]\epsilon_0\nabla\cdot\vec{E}=\rho+\rho_B[/tex]

[tex]\nabla\cdot\epsilon_0\vec{E}=\rho-\nabla\cdot\vec{P}[/tex]

[tex]\nabla\cdot(\epsilon_0\vec{E}+\vec{P})=\rho[/tex]

Now

Last edited:

- #5

- 406

- 6

D is a flux because it is a measurement of how much "charge" is "flowing" through a surface. Of course charges are actually moving, so I should really say it's a measure of the density of the electric field lines through that surface. Where the electric field lines bunch up, the electric flux density is high, and where they are spare the flux density is low. Of course, implicitly I mean this to be the case for surfaces that the field lines are perpendicualar to, but in vector notation D is considered apart from the surfaces you integrate it over.

- #6

- 1,306

- 19

If there is a charge enclosed by the surface of integration then you will get a net outward flux through the surface. The net outward flux is equal to the charge enclosed.

- #7

- 459

- 5

Thanks guys. Now I know.

BTW, is there something wrong with calling**D** the "electric field strength" (just like we call **H** the "magnetic field strength")?

BTW, is there something wrong with calling

Last edited:

- #8

- 905

- 4

- #9

- 406

- 6

E and H are fields. (V/m, A/m)

D and B are fluxes. (C/m^2, Wb/m^2)

D and B are fluxes. (C/m^2, Wb/m^2)

- #10

- 5,820

- 1,123

Thanks guys. Now I know.

BTW, is there something wrong with callingDthe "electric field strength" (just like we callHthe "magnetic field strength")?

In my opinion, one should use the "operational" units [or better, the arguably more fundamental differential-form] as a guide to naming these quantities.

Via differential forms [https://www.amazon.com/dp/0486655822/?tag=pfamazon01-20&tag=pfamazon01-20, etc...]:

- D, a [twisted] two-form with units Coulomb/m^2 [as mentioned above],

appears in Gauss' {Electric] Law and Ampere-Maxwell's Law

(some names: electric flux-density, electric induction, [di-]electric displacement, electric excitation)

- B, a two-form with units Weber/m^2,

appears in Gauss' [Magnetic] Law and in the Lorentz Force Law

(some names: magnetic flux-density, magnetic field induction, magnetic field strength*^{less common})

- E, a one-form with units of Volt/m,

appears in Faraday's Law and in the Lorentz Force Law

(some names: electric field strength, electric field intensity)

- H, a [twisted] one-form with units of Ampere/m,

appears in Ampere-Maxwell's Law

(some names: magnetic field strength, magnetic field intensity, magnetic excitation)

It seems that, historically, there has been some conflicting naming conventions regarding H and B.

Some groupings to note [as mentioned above]:

- flux-densities are associated with the 2-forms: D and B

field-intensities are associated with 1-forms: E and H - forces are associated with E and B

sources are associated with D and H (hence, "excitations") - upon making the usual [Hodge-dual] association with these differential-forms and vectors, E and D-vector are "polar vectors", whereas B and H are "axial vectors"

It would be nice if these names [including those from the more-standard textbooks (Jackson, etc...) ] were surveyed and all sorted out.

Last edited by a moderator:

- #11

marcusl

Science Advisor

Gold Member

- 2,762

- 412

Here's a short survey from an earlier thread

https://www.physicsforums.com/showthread.php?t=147184"

https://www.physicsforums.com/showthread.php?t=147184"

As for "getting it all sorted out", it didn't happen during the first 150 years of E&M theory. Maybe the next 150? :rofl:

https://www.physicsforums.com/showthread.php?t=147184"

More on whether to call H or B the magnetic field:.. because E and B are the fundamental field quantities, while D and H are derived [Jackson] "as a matter of convenience to take into account in an average way the contributions ... of atomic charges and currents." That is why E and B should be called fields, and why Mel Schwartz doesn't bother to even name H in his book. Furthermore, D is almost universally called Electric Displacement, and only rarely "dielectric flux density."

A survey of E&M books on my shelf shows B is called

Magnetic Induction -- by Smythe, Stratton, Jackson, Reitz & Milford

Magnetic Field -- Schwartz, Weber

alternately

Magnetic Flux Density -- Weber, Jackson

H is called

Magnetic Field or Field Intensity -- Smythe, Stratton, Jackson

Magnetic Intensity -- Weber, Reitz & Milford

unnamed -- Schwartz

https://www.physicsforums.com/showthread.php?t=147184"

As for "getting it all sorted out", it didn't happen during the first 150 years of E&M theory. Maybe the next 150? :rofl:

Last edited by a moderator:

- #12

- 459

- 5

BTW, robphy the way you describe E, H, B, & D as geometric objects with "one/two-form;" what branch of mathematics would you need to study in order to become familiar with these types of terminologies?

- #13

- 5,820

- 1,123

BTW, robphy the way you describe E, H, B, & D as geometric objects with "one/two-form;" what branch of mathematics would you need to study in order to become familiar with these types of terminologies?

tensor calculus and differential geometry.

consult the references I linked in my previous post

- #14

- 905

- 4

tensor calculus and differential geometry.

consult the references I linked in my previous post

:yuck:

I've had a fair amount of experience with both, and I can say with scientific certainty that tensor calculus is the devil!

Seriously though, thanks for the links. To be honest I never even paid attention to the units of D and H. Now I know why they're useful for finding the field inside a material.

- #15

Meir Achuz

Science Advisor

Homework Helper

Gold Member

- 3,529

- 114

Your units for D and B are for flux densities.E and H are fields. (V/m, A/m)

D and B are fluxes. (C/m^2, Wb/m^2)

- #16

- 2

- 0

[tex]\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}[/tex]

This would mean that determining “flux density”, which is the area density of flux, undoes the process of integrating the field over the surface area and is exactly the same as the “field”, just a different word.

If it were taken as given that [itex]\vec{D}=\varepsilon\vec{E}[/itex], then not only the [itex]4\pi[/itex] would cancel but the permittivity as well, leaving:[tex]\nabla\cdot\vec{D}=\rho[/tex]

My problem is that I don’t have anywhere near the maths ability or the motivation to understand how to get a polarisation vector from the dipole moments of the molecules of a dielectric and then the associated function for potential but in Jackson’s book, its got:[tex]\nabla\cdot\vec{E}=\frac{1}{\varepsilon_{0}}(\rho-\nabla\cdot\vec{P})[/tex]

and with

I think that - D is the net outward flux and
- Polarisation does reduce the effect of E and
- P does equal 0xeE (0 and xe being positive numbers)

[tex]\vec{D}=\varepsilon_{0}\vec{E}+\vec{P}[/tex]

when polarisation apparently reduces

On the other hand, Jackson says that “the fundamental fields are

I have the same problem in magnetostatics with magnetisation [tex]\vec{H}=\frac{1}{\mu_{0}}\vec{B}-\vec{M}[/tex]

when paramagnets and ferromagnets with positive susceptibility supposedly increase the field instead of reduce it. So my question is what is the physical meaning of vectors

- #17

Meir Achuz

Science Advisor

Homework Helper

Gold Member

- 3,529

- 114

You might do better with "Classical Electromagnetism" by Franklin which answers your question in chapter 6.So my question is what is the physical meaning of vectors,P,MandD? Also, what is the point of the permittivity number if it just cancels with that in the denominator of Coulomb’s constant when multiplying it withH? Someone please help show me where I’m going wrong.E

- #18

- 2

- 0

- #19

Meir Achuz

Science Advisor

Homework Helper

Gold Member

- 3,529

- 114

E,D,P,B,H,M. You just have to look at any text that uses gaussian units.

In gaussian units, E is the field due to all charges (free and bound).

D is the field due to only the free charges.

P is the dipole moment per unit volume of the matter.

-div P gives the bound charge density in the matter,

and P_n at the surface is bound surface charge.

If D=E+4pi P, then div D=4pi rho_{free}, and div E=4pi rho_{free+bound}

Share: