Talk:Electric displacement field/Archive 1

Displacement Current
There is a related page under Displacement current. --62.252.224.14 18:24, 11 Jun 2005 (UTC)


 * I'm not sure they should be merged. If so, it should be that Displacement current is merged here and then deleted, which means going through vfd for displacement current first.  It definitely shouldn't be the other way around though, so I'm not sure what you intended with the mergewith tag.  --Laura Scudder | Talk 05:58, 12 Jun 2005 (UTC)
 * Both of them are terms found in study of electromagnetism. In my opinion they should both be kept. That they are stub does not mean that they should be merged. It should mean that they should be expanded. AnyFile 15:21, 29 July 2005 (UTC)

i wanna know the main difference between electric field intensity and electric flux density?

Vector Notation
It appears there are two different notations for vectors on this page! One uses arrows, the other uses bold. This would be intensely confusing for someone new to these concepts, I think. Other electromagnetics pages (like electric field) seem to use bold. Although the arrows are particularly evocative, I think consistency would be better here. Unless anyone has a reason not to, I'm going to change this later. Xezlec 04:21, 31 December 2005 (UTC)


 * Ok, I've made everything bold. I also moved the general definition of D from the "interpretation" section to the definition at the top, since after all the D=$$\varepsilon$$E formula is really a special case and is not true for all materials everywhere.  While I was at it, I slightly changed the definition to mention bound charges and displacement current.  Displacement current, I think, is fairly important to understanding what D really is.  I have now stated that D exists to account for the effects of bound charges, as I was always taught.  If it has a broader scope than bound charges nowadays, I hope someone will fix it. Xezlec 00:39, 2 January 2006 (UTC)

Unless someone else gets to it first, I'm also going to at least add some reference to displacement current, and I'd like to explain a little further what flux density is, what the term comes from, what it represents, and what polarization is in this context. Xezlec 20:39, 31 December 2005 (UTC)

sdfasdfasdfsadfsad —Preceding unsigned comment added by 210.125.178.89 (talk) 13:18, 6 October 2008 (UTC)

Errors
The displacement field is not the macroscopic average of the electric field. All three of E, D, and P in D=E+P are macroscopic averages. Nor is D really the "generalization" of E. It is a distinct quantity, which happens to be useful in conjunction with E for describing media with bound charges.

See e.g. Jackson, Classical Electrodynamics, where a clear distinction is made between the macroscopic Maxwell's equations (which are almost universally what one uses in classical electromagnetism: all of the fields are averages) and the microscopic equations.

—Steven G. Johnson 19:39, 3 January 2006 (UTC)


 * The editor who added that is clearly attempting to summarize Griffiths's explanation of macroscopic fields in contrast to local microscopic fields, although poorly. Personally I think that whole section can go, otherwise, I'll just have to pull out my Griffiths when I get home and fix it.  &mdash; Laura Scudder &#9742; 19:46, 3 January 2006 (UTC)
 * You're both right, and since neither of you has touched it, I'm just going to delete the offending paragraph myself. :) Xezlec 20:43, 29 May 2006 (UTC)

Oh no!
Oh Jeez!--OK Stevie, Ive removed even more stuff. Is it OK now?--Light current 23:04, 2 June 2006 (UTC)


 * I've removed the disputed tag after a couple of more minor revisions. The article is obviously incomplete and somewhat messy, but the statements I was complaining about seem to be gone and I don't see any other obvious errors of fact.  I retitled the "Capacitor interpretation" section since this is more an application of D than an interpretation thereof.    —Steven G. Johnson 19:29, 4 June 2006 (UTC)

Dispute?
Is this article still disputed?

And to Light Current, I'd like to discuss the phrasing of the controversial first paragraph of "interpretation". To say the displacement field "is" the electric field, regardless of any modifying clauses that come afterward, seems a little off. By the logic of your phrasing, the electric field in free space would be synonymous with the displacement field. But that isn't true! They differ by a (not unitless) proportionality constant. In the (I think) older convention of referring to D and E using the same units and normalizing so that epsilon naught is one, that might make sense, but that isn't the convention used in this article. Xezlec 21:25, 29 May 2006 (UTC)
 * I didnt write it. I just modified the phrasing. Some one (you it appears) said 'can be thought of as'... Now it either is or it isnt! Please modify as you see fit. 8-)--Light current 22:27, 29 May 2006 (UTC)


 * Bull! What I said was "can be considered a way of thinking about", not just "can be considered as".  You changed "a way of thinking about" to "to be".  A way of thinking about something is quite different from the thing itself! Xezlec 17:37, 4 June 2006 (UTC)


 * Yep OK... over to you and Steven then! 8-|--Light current 18:40, 4 June 2006 (UTC)
 * The phrasing is still wrong. The displacement field can't "be thought of as the electric field after taking into account ..." or at least, this is misleading.  The displacement field and the electric field satisfy completely different boundary conditions, and in the static case the electric field is conservative (zero curl) while the displacement field is not.  The electric field has a direct interpretation in terms of the force on a charged particle, whereas the displacement field does not.  (The units are mostly irrelevant to the physical interpretation, by the way; it's just an arbitrary convention.)  —Steven G. Johnson 22:12, 29 May 2006 (UTC)


 * Steven, please see my reply to above post. It appears User:Xezlec changed the para meaning most recently- not me-- honest!--Light current 22:27, 29 May 2006 (UTC)


 * That's just not true! Look in the history.  My last change was 15:01, May 26, three days before your comment right here.  You then changed it on 15:06 (five minutes later), and only you and H2g2bob have touched it since then.  Is there some kind of glitch in the system?  Xezlec 17:37, 4 June 2006 (UTC)


 * It doesn't matter to me who is responsible for the phrasing. —Steven G. Johnson 19:00, 30 May 2006 (UTC)
 * Any way it doesnt really matter. Ill leave you and Steven J to sort it out 8-)--Light current 18:42, 4 June 2006 (UTC)


 * Well why dont you just change it? 8-)--Light current 19:48, 30 May 2006 (UTC)

Just to address the point "But that isn't true! They differ by a (not unitless) proportionality constant." - In my opinion this is irrelevant as quatities (e.g. fields) can be expressed in different units, but this doesn't change their nature - e.g. in particle physics one often measures rest mass in electron volts. To convert this to kilograms requires a not unitles proportionality constant but it is surely wrong to suggest that the mass expressed in one way is different from the mass expressed in another. --Neo 09:46, 30 May 2006 (UTC)


 * Apologies. I guess I meant "not dimensionless".  Is that better? Xezlec 17:39, 4 June 2006 (UTC)


 * The first paragraph looks ok to me now (it was edited by Light current). If there are no objections, I'd say that this should no longer be tagged as disputed. --H2g2bob 20:48, 1 June 2006 (UTC)


 * Well I dont think its disputed really--Light current 22:11, 1 June 2006 (UTC)


 * Except that it is still wrong. D does not "generalize" E.  It is a distinct quantity. —Steven G. Johnson 03:05, 2 June 2006 (UTC)


 * Yes it is a distinct quantity, but arguably it is a distinct quantity whose reason for existing in our repetoire of mathematical ideas is to generalize the concept of an electric field. Well, okay, it's debatable. Xezlec 17:37, 4 June 2006 (UTC)


 * Is that better now Steven? Are you happy for the disputed tag to come off now?--Light current 08:17, 2 June 2006 (UTC)


 * Nope: "The electric displacement field can be considered to be the electric field after taking into account..." is still wrong, as I explained above. (Sorry I'm just griping from the sidelines, but I really don't have time to rewrite the article myself.) —Steven G. Johnson 21:23, 2 June 2006 (UTC)


 * And that is exactly what I've been saying. And the edit I made (on 15:01, May 29), which Light Current immediately undid five minutes later, fixed that problem. Xezlec 17:37, 4 June 2006 (UTC)


 * I apologize if I unfixed your fix. This was not not intentional. Yes looking back on it I can see that was a sloppy edit by me 8-(. I was just trying to fix the weasle wording but in the exceitement, I threw the baby out as well. Aplogies to all concerned! Anyway Ill keep out of it until you 2 have settled!--Light current 18:45, 4 June 2006 (UTC)

The displacement field clearly is the complete (and therefore "true") electric field. It so happens that it equals $$\varepsilon_0 \mathbf{E}$$ when relative permittivity is 1, which is what you expect out of electric fields in an vacuum. The relationship of D (coulombs/meter^2) to E (volts/meter) is the same as the relationship of B (webers/meter^2) to H (amps/meter). E ignores the dielectric polarization P that may or may not be a part of D. H ignores the magnetization M that may or may not be a part of B. It is also natural that the Poynting vector is defined as S=ExH (The Abraham version) or S=(DxB)c^2 (The Minkowski version, which is greater by n^2; See Abraham–Minkowski controversy). Basically the Abraham version ignores modes embedded in a material and is associated with the kinetic momentum, while the Minkowski version includes modes embedded in a material (phase space) and is associated with the canonical momentum. One can also use an ExB form for electric monopole current (again, E ignores dielectric polarization) which makes sense as the Electric scalar potential (voltage = energy / charge) and Magnetic vector potential (webers/meter = momentum / charge) can be used to express in the E and B fields, with the Magnetic vector potential basically representing a medium of electrical currents whose vorticity corresponds to the B-field; Note that the potentials overlook the effects of electric permittivity but not those of magnetic permeability. The P-field is completely left out of the potentials, and therefore so is the D-field! Alternatively, one can use a DxH form for magnetic monopole current, but again, H ignores magnetization. I recommended reading the following paper (http://arxiv.org/pdf/0908.1721v2.pdf).siNkarma86—Expert Sectioneer of Wikipedia undefined 09:10, 11 April 2013 (UTC)

Displacement field in a capacitor
I have problems with the current example:

''Consider an infinite parallel plate capacitor placed in space (or in a medium) with no free charges present except on the capacitor. In SI units, the charge density on the plates is equal to the value of the D field between the plates. This follows directly from Gauss's law, by integrating over a small rectangular box straddling the plate of the capacitor:'' [...]


 * 1) a plate capacitor has two plates. Therefore is the plate considered to straddle both plates or only one?
 * 2) Is Q the charge of only one plate or of both?
 * 3) What size is A in $$|\mathbf{D}| = \frac{Q}{A}$$ supposed to be when we are talking about a infinite parallel plate capacitor?


 * I think the current version in the article is wrong - the rectangular box should only straddle one plate.
 * The integral is taken over all sides of the box. As stated in the article, the sides that are perpendicular to the plates have no component of the field through them so they do not contribute to the integral. That leaves the two parallel sides.
 * Outside the plates there is no field (due to the effects from each plate cancelling) - if we consider a box that straddles both plates, then each of the parallel sides lies outside the plates and the whole integral evaluates to zero. We must hence consider a box that straddles just one plate. We then have a side of area A (lying in between the plates) being cut by the field. Since the field is perpendicular to the side, the vectors in the integral become scalars and the LHS = |D|A. The RHS gives the amount of charge stored on the capacitor plate over a certain area A. Dividing through by A then gives the required result that the magnitude of the field is equal to the charge per unit area (charge density) on one plate of the capacitor.


 * In answer to your question, the size of A is not important - it is simply an arbitrary area on which a charge Q is found on one plate of the capacitor. We are assuming that the charge is uniformly distributed across the plates. We must consider an infinite parallel plate capacitor so that we can neglect end effects at the edges of the capacitor. 129.169.173.182 (talk) 11:06, 22 August 2008 (UTC)

Thanks, --Abdull 10:49, 22 October 2007 (UTC)

Clarity
I think that there should be some additional language to define the (lack of) physical meaning of the term, such as can be found at http://www.physicsforums.com/library.php?do=view_item&itemid=6. As is pointed out there, "Technically, only the polarization field, P, is a displacement field." Rasraster 22:15, 5 November 2008 (UTC) —Preceding unsigned comment added by Rasraster (talk • contribs)

definition is backwards
I was taught that the relation between D and E is defined by :$$\mathbf{D} = \mathbf{E}+4 \pi \mathbf{P} $$ (may be slightly different in SI units). Only in special circumstances can a linear relationship between D and E be assumed, as in $$\mathbf{D} = \varepsilon \mathbf{E}$$. If the latter were always true, there would be no need to even invoke D in Maxwell's equations since you could just write $$\varepsilon \mathbf{E}$$ instead. Is my interpretation wrong or does the article need changed? misli h  19:32, 10 June 2009 (UTC)
 * I think you're correct about the definition of displacement current. Certainly, this is how the textbooks that I've read define it.  Please go ahead and correct it.  Thanks! —老陳 (留言) 13:38, 11 June 2009 (UTC)

Diagram
Do we need a diagram such as :




 * No, I don't think so... displacement fields aren't really that  closely associated with capacitors and circuit diagrams, even in pedagogy (whereas displacement currents are another matter), so that's probably not the best direction to take the article in.  —Steven G. Johnson 21:59, 5 June 2006 (UTC)

OK!--Light current 22:04, 5 June 2006 (UTC)

Same Thing as Electric Flux Density?
Isn't the electric displacement field the same thing as Electric Flux Density? See: http://www.globalscience24.com/eng/d/electric-displacement-electric-flux-density/electric-displacement-electric-flux-density.htm Aaron Myles Landwehr (talk) 16:34, 26 September 2009 (UTC)