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Electric displacement field across dielectrics

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Zorba
#1
Dec20-10, 03:38 PM
P: 77
So suppose we have two dielectrics in contact and we want to know how [tex]\mathbf{D}[/tex] varies across them, then we can use the fact that since [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] and we have no "free" charges at the boundary then [tex]\mathbf{D}[/tex] is continuous across it.

So my question is, in Grant & Philips they seem to suggest that only the components of [tex]\mathbf{D}_{1,2}[/tex] that are perpendicular to the surface of contact of the dielectrics, are equal. Is it the case that the components parallel are not also equal? And if so why, because [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] seems to imply that they should be equal?
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yungman
#2
Dec20-10, 06:49 PM
P: 3,898
Quote Quote by Zorba View Post
So suppose we have two dielectrics in contact and we want to know how [tex]\mathbf{D}[/tex] varies across them, then we can use the fact that since [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] and we have no "free" charges at the boundary then [tex]\mathbf{D}[/tex] is continuous across it.

So my question is, in Grant & Philips they seem to suggest that only the components of [tex]\mathbf{D}_{1,2}[/tex] that are perpendicular to the surface of contact of the dielectrics, are equal. Is it the case that the components parallel are not also equal? And if so why, because [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] seems to imply that they should be equal?
Only the component of D normal to surface is continuous cross boundary if no surface charge present.

Read the boundary conditions, use integrated form:

[tex]\int_V \nabla \cdot \vec D dV = \int_S \vec D \cdot \hat n dS \hbox { where } \hat n \hbox { is the unit normal of the surface.}[/tex]

[tex]\int_V \nabla \cdot \vec D dV = \int_S \vec D \cdot \hat n dS = \rho_V[/tex]

[tex] \vec D \cdot \hat n = D_n[/tex]

Therefore only the normal component of D is continuous cross boundary.
Zorba
#3
Dec21-10, 09:11 AM
P: 77
Okay, and if we draw a closed surface that encompasses a part of both of the materials and crosses the boundary, with no free charges we have [tex]\int_S \vec D \cdot \hat n dS=0[/tex], so then the total flux out of the surface is zero, so that would imply that D_{1,2} are just equal, not just the components normal to the surface no?

yungman
#4
Dec22-10, 01:58 AM
P: 3,898
Electric displacement field across dielectrics

Quote Quote by Zorba View Post
Okay, and if we draw a closed surface that encompasses a part of both of the materials and crosses the boundary, with no free charges we have [tex]\int_S \vec D \cdot \hat n dS=0[/tex], so then the total flux out of the surface is zero, so that would imply that D_{1,2} are just equal, not just the components normal to the surface no?
I am no expert, I just want to go through this with you and see what happen.

My understanding is you draw a closed 3D surface S1 that enclose part of the dielectric. Since there is no free change so:

[tex]\int_{S_1} \vec D \cdot \hat n dS_1=0[/tex]

What this mean is the surface S1 you draw have no net flux exit from the closed surface. There are flux coming out of the dielectric into the closed surface S1 that you defined. All it said is the total flux going into the surface is equal the the total flux exit the surface so the net flux out of the surface S1 is zero. That is all the equation said, nothing more.

Put it in another light, there are flux coming out of the dielectric surface that is exclosed by S1 that you defined, but it just go right out of the surface. So the net flux out of S1 is zero. This equation say nothing about the composition of [itex]\vec D_{21}[/itex] that you are talking about.

BTW I use [tex] \vec D_{21} \hbox { instead of } \vec D_{12} [/tex] because the flux you are talking is coming out from the dielectric into the space.

This is my understanding of this Maxwell's equation.
Born2bwire
#5
Dec22-10, 03:04 AM
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Quote Quote by Zorba View Post
Okay, and if we draw a closed surface that encompasses a part of both of the materials and crosses the boundary, with no free charges we have [tex]\int_S \vec D \cdot \hat n dS=0[/tex], so then the total flux out of the surface is zero, so that would imply that D_{1,2} are just equal, not just the components normal to the surface no?
Take your Gaussian surface down to an infinitesimal volume centered about the boundary.

For example, let's just assume we have a rectangular box that runs through the boundary. As we make the box smaller along the dimensions parallel to the boundary we begin to approach the limit that the same field lines that enter the sides are the same field lines that exit out the opposite sides (The sides normal to the boundary, these would be the field lines tangential to the boundary). Nothing special there since that behavior would occur regardless of the boundary.

So let's start over and instead squeeze the box so that its height approaches zero and it becomes flush with the boundary. Now what happens? The flux passing through the bottom side is due to the fields normal to the boundary on the side in medium one. The flux passing through the upper side is due to the fields normal to the boundary on the side in medium two. Now Gauss' Law says that the flux through this volume is equal to the enclosed charge. Since we squeezed the box so that it has an infinitesimal height, there isn't any flux coming in the sides and the total volume is equal to the area that the box is enclosing on the boundary. So assuming a uniform charge distribution on the boundary then we know that

[tex] D_{1n}-D_{2n} = \rho_f [/tex]

So in a source free region there is no free charge and thus the normal component of the electric flux density is zero. Likewise, we can use Faraday's Law in a similar manner to find that the tangential electric field has to be continuous (not the electric flux density though).
yungman
#6
Dec22-10, 03:41 AM
P: 3,898
Quote Quote by Born2bwire View Post
Take your Gaussian surface down to an infinitesimal volume centered about the boundary.

For example, let's just assume we have a rectangular box that runs through the boundary. As we make the box smaller along the dimensions parallel to the boundary we begin to approach the limit that the same field lines that enter the sides are the same field lines that exit out the opposite sides (The sides normal to the boundary, these would be the field lines tangential to the boundary). Nothing special there since that behavior would occur regardless of the boundary.

So let's start over and instead squeeze the box so that its height approaches zero and it becomes flush with the boundary. Now what happens? The flux passing through the bottom side is due to the fields normal to the boundary on the side in medium one. The flux passing through the upper side is due to the fields normal to the boundary on the side in medium two. Now Gauss' Law says that the flux through this volume is equal to the enclosed charge. Since we squeezed the box so that it has an infinitesimal height, there isn't any flux coming in the sides and the total volume is equal to the area that the box is enclosing on the boundary. So assuming a uniform charge distribution on the boundary then we know that

[tex] D_{1n}-D_{2n} = \rho_f [/tex]

So in a source free region there is no free charge and thus the normal component of the electric flux density is zero. Likewise, we can use Faraday's Law in a similar manner to find that the tangential electric field has to be continuous (not the electric flux density though).
I thought [tex]\int_{S_1} \vec D \cdot \hat n d S_1=0[/tex] has nothing to do with whether the [tex]\vec D[/itex] is continuous cross boundary. The equation only say if there is no free charge inside the S1. Therefore NET flux in and out of S1 is zero. It really don't care about whether D is continuous cross boundary of the dielectric.

In another word, using the gauss surface S1 prove nothing about the continuity of the D at the boundary of the dielectric. Why do we need to concern with the boundary condition whether the normal or the tangential component is continuous? The two should be like apple and orange, just courious.
Born2bwire
#7
Dec22-10, 03:52 AM
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Quote Quote by yungman View Post
I thought [tex]\int_{S_1} \vec D \cdot \hat n d S_1=0[/tex] has nothing to do with whether the [tex]\vec D[/itex] is continuous cross boundary. The equation only say if there is no free charge inside the S1. Therefore NET flux in and out of S1 is zero. It really don't care about whether D is continuous cross boundary of the dielectric.

In another word, using the gauss surface S1 prove nothing about the continuity of the D at the boundary of the dielectric. Why do we need to concern with the boundary condition whether the normal or the tangential component is continuous? The two should be like apple and orange, just courious.
You can use the integral form of Gauss' Law to derive the boundary condition that the normal component of the electric flux density is continuous across a boundary (when no free charges are present, we can get the general case this way too). You can then use the integral form of Faraday's Law to derive the boundary condition that the tangential electric field is continuous. With these two we have the boundary conditions necessary to deal with inhomogeneous media. One can also derive the same boundary conditions by judicious analysis of the ordinary differential equations that arise when solving for TE and TM plane waves in planarly layered media.

To derive this via the integrals, it's just a matter of taking the Guassian surface to the limit of approaching the boundary itself. When doing this then the flux is simply the difference between the field incident on the two sides of the boundary. When deriving it from the ordinary differential equations the boundary conditions fall out of the requirement that we have finite terms in the ODE.
Zorba
#8
Dec22-10, 05:12 AM
P: 77
@Born2Bwire
Thanks, that's the same argument they use in Grant & Philips, I get all of that, my question though is regards that argument, it seems arbitrary that we state that the normal components are equal, why can we not just apply the same arguments to a box where the height tends to zero and show that the other components are equal, because as far as I can see the div D = rho_f argument does not distinguish between two separate materials, viz. if we applied the surface argument to just one material or two, then the equations stay the same there is no inherent distinction in their form...

This is getting frustrating, as I don't think I can articulate what I mean properly, hopefully that'll give you an idea though.
Born2bwire
#9
Dec22-10, 05:35 AM
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But remember that the D field incorporates the material information. The primitive here is the electric field. The electric flux density is the summation of the electric field and the polarization of the media.

As I mentioned above, if you change the process so that you now make the box have an infinitesimal width along the plane normal to the boundary, then we do not gain any new information. Because now the field lines are tangential to the boundary. Before, the distinction came from the fact that the field lines on one side of the box was in medium 1 and the other was in medium 2. But if we were to rotate the problem then the field lines on one side of the box are in the same medium when they exit the box. So there is nothing new that we gain here.

If we want to look at the tangential boundary conditions we use Faraday's Law.
Zorba
#10
Dec22-10, 12:07 PM
P: 77
Quote Quote by Born2bwire View Post
But remember that the D field incorporates the material information. The primitive here is the electric field. The electric flux density is the summation of the electric field and the polarization of the media.
Are yes, that's what I was missing, the permittivity in the expression for D.

There's still a few things that make me feel I'm missing some stuff here, I think I'll have to go and think a bit about what's physically going in with the divergence of a field and so on to try and clear thinks up. Thanks for the help.
asmani
#11
Jan3-11, 11:01 AM
P: 86
Quote Quote by Zorba View Post
So suppose we have two dielectrics in contact and we want to know how [tex]\mathbf{D}[/tex] varies across them, then we can use the fact that since [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] and we have no "free" charges at the boundary then [tex]\mathbf{D}[/tex] is continuous across it.

So my question is, in Grant & Philips they seem to suggest that only the components of [tex]\mathbf{D}_{1,2}[/tex] that are perpendicular to the surface of contact of the dielectrics, are equal. Is it the case that the components parallel are not also equal? And if so why, because [tex]\nabla \cdot \mathbf{D}[/tex] seems to imply that they should be equal?
I think the key is in the Helmholtz's theorem; you need both [tex]\nabla \times \mathbf{D}[/tex] and [tex]\nabla \cdot \mathbf{D}[/tex] to determine whether D is independent of dielectrics or not.
The [tex]\nabla \times \mathbf{D}[/tex] depends on dielectrics; it's not zero at boundary of dielectrics.
I'm not sure even if the equation [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] is valid at boundary of dielectrics.
Born2bwire
#12
Jan3-11, 09:13 PM
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Quote Quote by asmani View Post
I think the key is in the Helmholtz's theorem; you need both [tex]\nabla \times \mathbf{D}[/tex] and [tex]\nabla \cdot \mathbf{D}[/tex] to determine whether D is independent of dielectrics or not.
The [tex]\nabla \times \mathbf{D}[/tex] depends on dielectrics; it's not zero at boundary of dielectrics.
I'm not sure even if the equation [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] is valid at boundary of dielectrics.
It is valid at the boundary, it's one of the boundary conditions that ensure uniqueness of the solutions to Maxwell's Eqations. The derivation was described earlier in this thread and can be found in any undergraduate electromagnetic textbook.
asmani
#13
Jan4-11, 02:36 AM
P: 86
Quote Quote by Born2bwire View Post
It is valid at the boundary, it's one of the boundary conditions that ensure uniqueness of the solutions to Maxwell's Eqations. The derivation was described earlier in this thread and can be found in any undergraduate electromagnetic textbook.
You are of course right, I have seen the derivation in Cheng's book.
In this book, two equivalent charge distribution are introduced for dielectrics, one is a volume density [tex] \rho_p=-\nabla \cdot \mathbf{P}[/tex] and another is a surface density [tex] \rho _{ps}=P\cdot a_n [/tex].
In derivation of the equation [tex] \nabla \cdot \mathbf{D} = \rho_f [/tex], only the volume density is entered to the calculations and the surface density seems to be ignored.
Then in derivation of boundary conditions of D, it says [tex]D_{1n}-D_{2n}=\rho _s[/tex].
I thought maybe [tex]\rho _s[/tex] is the same surface density which seemed to be ignored.

What about the curl? Is this derivation valid?:

[tex] \nabla \times \mathbf{D}=\nabla \times \mathbf{(\epsilon E)=\epsilon \nabla \times \mathbf{E} -E\times \nabla \mathbf\epsilon }[/tex]
Since:
[tex]\nabla \times \mathbf{E}=0\rightarrow \nabla \times \mathbf{D}=-E\times \nabla \mathbf\epsilon [/tex]

This means that [tex] \\\nabla \times \mathbf{D}[/tex] could be nonzero and depends on dielectrics.

Thanks a lot
Meir Achuz
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Jan4-11, 04:15 PM
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Quote Quote by Zorba View Post
So suppose we have two dielectrics in contact and we want to know how [tex]\mathbf{D}[/tex] varies across them, then we can use the fact that since [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] and we have no "free" charges at the boundary then [tex]\mathbf{D}[/tex] is continuous across it.

So my question is, in Grant & Philips they seem to suggest that only the components of [tex]\mathbf{D}_{1,2}[/tex] that are perpendicular to the surface of contact of the dielectrics, are equal. Is it the case that the components parallel are not also equal? And if so why, because [tex]\nabla \cdot \mathbf{D} = \rho_f[/tex] seems to imply that they should be equal?
E_T (tangential) is continuous, so D_T/epsilon is the same on each side and D_T must change.


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