Electric field boundary equation implication at air/earth interface

  • Thread starter Stanley514
  • Start date
  • #1
411
2
The greatest problem of thermoelectrics is the need to maintain very low thermal conductivity.
How is it possible that pyroelectrics do not have this limitation and do not need temperature differences to produce electricity from heat?If we will heat all pyroelectric body uniformly it will still generate electricity,right?I know that pyroelectrics work in impulse maner and should be heated in impulses.Therefore pyroelectrics could be more efficient than thermoelectrics at least in cascade.
Disadvantage of pyroelectrics is that they produce very high voltage.This trait could be used even for nuclear fusion purposes.I thought that it would be valuable to create some material which would have best traits of both.Ability to work in impulses and produce low voltage and high current.
What will happen if we will heat thermoelectric in impulse mode?
 

Answers and Replies

  • #2
65
0
Hello folks.
I'm a little confused due to a problem.
Presume there are two plates which are perpendicular to each other, just like a plus symbol.

One of these plates has positive charge density, and the other one has negative charge density.

No information whether they are insulating or conducting, but they touch each other.

Now if we would like to draw the electric field lines. How it looks like ?

I thought that since these plates in contact, they become neutral and don't create any electric field, so no lines.

However my teacher said that, since they go to infinite, it takes infinite time to get neutral, so there must be electric field lines from each positive ends to negative ends.

I'm not satisfied with that answer. What do you think ?
 

Attachments

  • #3
1
0
Hi all,
I'm trying to find which magnetic materials can operate without too much losses at about 2.45 GHz and have little relative magnetic permeability, i.e., in the range 15-35.

About the same for dielectric materials, with a relative electric permittivity in the range 2-7.

Moreover, I ask if a magnetic material designed as a substrate for a planar antenna (cylindrical shape) may be used without having been magnetized first. I think the only difference in using it magnetized near the remanence region would be in relative permeability.

Thanks for the reply
 
  • #4
370
0
I've recently bought myself a tabletop Van De Graaff generator and have been testing the effect of various objects on the field it produces.

When I hold a small polystyrene ball on a string near the electrode I find that the ball is pulled strongly towards it.

Apparently this is due to the ball being polarized by the field.

Now I tried two experiments.

1/ I held a metal tray between the electrode and the polystyrene ball. The ball just hung down on the thread indicating that the metal tray had blocked the field. This is what I would have expected.

2/ Next I held a single piece of paper between the electrode and the polystyrene ball. Again the ball hung down vertically on the thread indicating that the paper had blocked the field.

Would you expect a piece of paper to block the electric field from a 100 kV electrode?

Is this effect due to the paper's dielectric properties?

John

PS I checked that the metal tray and the paper was not discharging the electrode by holding them near to the electrode but away from the ball-on-a-string. The ball's attraction to the electrode in this case was not affected.
 
  • #5
2
0
I have a question concerning the electric field of a rotating linear dielectric disk (with angular velocity w) and a static B-field that is directed parallel to its normal (it is in the z-direction if we use cylindrical coordinates).

Basically, how do I obtain the electric field. I understand that if the disk were metal, then the total force would be zero inside, and therefore the radial electric field would be given by the Lorentz force law.
But for this problem, Is there is force in the 'meat' of the dielectric?
I think the Lorentz force, while in the rotating frame, is the electric field plus the force from a moving charge in a magnetic field (F = qE + qvB(1/c) ) with v = wr. I could start from the polarizability, but that is linearly proportional to the total electric field inside. I'm confused here. All help is appreciated!
 
  • #6
25
0
Code:
Suppose the dielectric material is fixed in position and filling the capacitor, and you would have this term in the way of calculating something.

[tex]\int\nabla\cdot\left[\left(\Delta{D}\right){V}\right]{d}\tau[/tex]

where D is the dielectric displacement

Now that turns into (by divergence theorem):

[tex]\int\Delta\left({D}{V}\right)\cdot{d}a[/tex]

Now my textbook says that this term would vanish if we integrate over all of space. Why is that? Thanks for your help in advance. =)[/QUOTE]
 
  • #7
1
0
Hi all,

In essence I'm trying to differentiate between the "perfect conductor" approximation in electrostatics, and a real metal.

I'm specifically trying to determine the surface charge distribution at the top and bottom of a slab of metal (any will do) when it's inserted between the plates of a parallel plate capacitor (say it's 1/3 the size of the gap between the plates, centered between them) and +- V is applied to the upper and lower (perfectly conducting) plates respectively. Let the rest of the space between the plates have permittivity eps_r.

I know the positive and negative voltages should draw charge to the surface of the metal. Originally, the metal may have a uniform distribution of electrons at density n_0. With the applied voltage I'm thinking it should separate to a net negative charge at the upper surface and net negative at the lower surface. This yeilds a voltage drop across the metal. This, however, is in contrast to setting the metal as a perfect conductor, in which case, there is no voltage drop.

Can anyone point me in the direction to first, the discrepancy in 'real metal' voltage drop vs. 'perfect conductor' zero voltage drop.
And, second, how you could determine the distribution of charge (perpendicular to the metal upper and lower faces), which my instinct says should be linear?

Ok, thanks for any input you can give me!

--chewy
 
  • #8
I have a proposed solution to the following problem, but I don't know if is correct, and so I'll ask you for help:

A thin dielectric sheet of uniform charge density [itex]\sigma[/itex] (I'll assume [itex]\sigma > 0[/itex] for simplicity) is surrounded by two thin conductors, which are initially uncharged and are connected by a wire. The first and second conductors are at distances [itex]a, b[/itex] respectively from the dielectric. The three sheets are parallel. The configuration is then like a sandwich. I want to find the electric field in every point of the space.

My proposed solution is that the electric field between the conductor A and the dielectric, and between the conductor B and the dielectric, is null. The electric field is [itex]|\bar{E}_{outside}| = \dfrac{\sigma}{2\epsilon_0}[/itex] everywhere else.

My argumentation is as follows: for the electric field outside the sandwich, the solution is clear and comes from Gauss's law.

For the electric field inside, let's suppose that a charge [itex]\eta[/itex] is induced on the sheet A (and thus a charge [itex]-\eta[/itex] on B). This means that the electric field in the space between A and the dielectric will be [itex]E_{A} = \dfrac{\sigma+\eta}{2\epsilon_0}[/itex], while [itex]E_{B} = \dfrac{\sigma-\eta}{2\epsilon_0}[/itex]. The potential difference between A and B is then [itex]\Delta V = E_{B} b - E_{A} a[/itex].

But the hypothesis is that [itex]\Delta V = 0[/itex] (they are connected with a wire), and so the difference above only holds is [itex]\eta = \sigma[/itex], which means that [itex]E_{A} = E_{B} = 0[/itex].
 
  • #9
14
0
I am trying to understand how to deal with a boundary value problem when there are multiple dielectrics inside the volume.


I'll start with the classic question, then add the dielectric complexity.

We want to solve for the potential inside a charge-free axis-aligned 3D rectilinear volume.
5 of the 6 sides of the volume are at 0 potential. The sixth side (call it the bottom, with Z=0) has a known fixed potential V(x, y, 0). Solve for V(x, y, z) inside the volume.

This is a classic boundary value problem, and nearly every electrostatics text shows the solution by separation of variables, building up the base potential by a Fourier series, with each term decaying in Z based on its frequency.

Now to involve dielectrics. Take the same problem, but now the volume X<C is filled with an ideal linear dielectric with permittivity [itex]\epsilon_A[/itex] and the volume X>C is filled filled with an ideal linear dielectric with permittivity [itex]\epsilon_B[/itex].

What is the potential everywhere in the volume?


I see two ways of attacking this. My first thought is to take the original, known, solution for constant [itex]\epsilon[/itex] and look at the plane V(x=c, y, z), specifically at the Ex component of it (just [itex]\frac{\partial V}{\partial x}[/itex]). From dielectrics, we know the real solution is continuous in V here but discontinuous in Ex, so we can determine the amount of induced charge needed so it creates the required Ex ratio. Then use that induced charge to solve two more boundary value problem for each of the two halves independently, this time dealing with constant [itex]\epsilon[/itex]. Sum the three solutions for the net solution. This approach feels ugly, and I'm sure the algebra will explode into many lines of fiddly nested integrals. And I'm not even sure if it's a way to actually make the proper answer.

The other way to attack the problem is by images. All the electrodynamics texts talk about solving planar dielectric boundary value problems like this (for example, in Jackson, page 155, and Jackson, page 189) so it's natural to think that maybe V(x, y, z) could be represented by some weighted sum of the known constant [itex]\epsilon[/itex] solution, like A*V(x, y, z) + B*V(2C-x, y, z) for computable weights A and B (which would be different for the two [itex]\epsilon[/itex] regions.)

The image approach is a lot more elegant in that it is easy to evaluate, no pages of algebra, but it also can't be correct, because what if the image point is reflected across X=C to a point that's outside the bounding region? That makes no sense, so this simple guess can't be right.

But then maybe the images aren't using the same bounding region.. maybe for the left side, we reflect the left part of the bounding region itself over X=C to make it symmetric, solve that.. similarly for the right side, then compute some weighted sum of the two solutions... but now I'm just making up strategies with no justification. But as Griffith himself says on page 190, "like all image solutions, this owes its justification to the fact that it works... image solutions are always a matter of guesswork."

So, I open the question to the experts: what's the best way to attack this dielectric boundary value problem?

Thanks! It's a fun puzzle I think!
 
  • #10
292
1
In the ferroelectric literature, thermodynamic potentials are often written in terms of electric polarization P rather than displacement field D. When written in terms of D, the minimization of a thermo potential retains its equivalence to the entropy maximum principle, so there is a clear downside to using P. But aside from this, how does one construct the potentials with P? In the literature that I have found so far, it often looks like P is simply substituted for D, and the point is not explicitly discussed.

Secondly, according to the textbook by Lines and Glass, the other downside to using P in the thermodynamic potentials is that constant P conditions are not obtainable in practice. I am having trouble seeing how one obtains constant E or constant D conditions experimentally. The book makes some reference to open circuit conditions being constant D and short circuit being constant E. For short circuit, the field in the ferroelectric should be zero, and thus constant. Is it possible to obtain constant E conditions for E=\=0? For open circuit, I don't see how this corresponds to constant D at all.
 
  • #11
82
0
In a post I made in the materials engineering sub forum someone mentioned the field extends only a few atomic spacings beyond the surface of a crystal, in the case where I was thinking a micro crystal grain.

After talking to a physics lecturer he mentioned that due to the positive and negative charges from atoms and electrons that it is very plausible a field would be negated outside the bulk of the crystal.

Wikipedia mentions that a P vector is varied by the surface charge density, but this is not the source of the piezoelectric effect but that the bulks dipole density is.

Is there a type of piezo electric crystal that could have a field extend beyond the surface. And is this the case in electric lighters?

There is some mention of poling in the wiki article, alignment of the weiss domains that form around dipoles.
 
  • #12
since, B=curl(A), curl(E)= -[itex]\partial[/itex]B/[itex]\partial[/itex]t
1)curl(E)=- [itex]\partial[/itex]/[itex]\partial[/itex]t(curl(A))
2)curl( E+[itex]\partial[/itex]A/[itex]\partial[/itex]t)=0
3)then since curl([itex]\nabla[/itex]V)=0,

E +[itex]\partial[/itex]A/[itex]\partial[/itex]t =- [itex]\nabla[/itex]V
E= -[itex]\nabla[/itex]V -[itex]\partial[/itex]A/[itex]\partial[/itex]t


I'm confused about how to go from step 1 to step 2. The first thing I did was add the right side to the left to get: curl(E)+[itex]\partial[/itex]/[itex]\partial[/itex]t(curl(A))=0

I know there's a property that says
a X (b+c) = a X b + a X c

But what about the -[itex]\partial[/itex]/[itex]\partial[/itex]t ? How would I deal with that?

Thank you.

edit: perhaps this should be in the calculus section?
 
Last edited:
  • #13
1
0
Hello physics geniuses,

I am wondering if anyone can point me in the direction of a paper or book that describes a nonlinear optics equivalent of the Fresnel equations.

The Fresnel equations make boundary condition assumptions which do not account for higher order (nonlinear) terms in the electric susceptibility. There won't be an analytical solution to this, but some numerical simulations will do with some plots, at least to give an experimentalist some idea of how reflection behaves as a function of intensity, angle of incidence, p- or s- polarization, and refractive index.

If the above cannot be found, a normal-incidence-only version would be a start and still useful.

Thanks!!
 
  • #14
41
0
I've been reading up on methods of generating plasma via electrical discharges and I have some question. I'm learning this on my own here so please correct any faulty asumptions. It appears to me that a typical RF discharge across a dielectric cell operates below breakdown voltage and the plasma is generated via the Townsend avalanche. At some point (quickly) a sheath develops and the field in the plasma drops to near neutral although current still passes across the gap. Please correct me if I'm wrong.

Question 1: What is the advantage of using a dielectric material? The only thing I can think of is that the electrodes need protected to avoid some type of sputtering.

Question 2: I undestand the electon avalanche effect, but how does a current pass through the gap if the voltage is below breakdown. Does the applied field push the plasma electrons into the lower potential electrode and a higher energy electron busts through the sheath to take it's place? If so, wouldn't the electron temperature keep going up?

On the other hand, I've seen short duration HV discharges that fire well above breakdown. In this case the high potential pushes electrons forward causing ionization/breakdown, no? A sheath layer forms soon after, dropping the field in the plasma.

Question 3: Since the potential is higher, does that mean the sheath layer is much thicker? Also, how does/doesn't sheath formation coincide with breakdown in time? I'm guessing that most applied voltage waveforms much faster than sheath formation, so in this case breakdown would occur long before sheath formation? In either case, the field in the gap would drop, but I'm unclear on the two mechanisms.

Question 4: Are there analytical solutions that account for both of these processes at once? Neglecting inelastic collisions by electrons for the moment, the diffusion processes could be modeled to show the evolution of breakdown and sheath formation? If so, have they been done in multiple dimensions (for something other than infinite planar electrodes).
 
  • #15
411
2
There is such an issue in dielectric capacitors as dielectric absorption.Usually it is regarded as drawback.I have some questions on it: 1)Does dielectric absorption allow you to store more charge at the same voltage?And therefore materials with the same dielectric constant and dielectric strength but higher dielectric absorption have larger energy density?If yes what is mechanism of it?Or higher dielectric absorption means higher voltage?
2)What is relation between dielectric absorption in dielectrics and space-charge polarisation?Is it allways the same thing?
 
  • #16
1
0
Hey there, I've been reading a lot about piezoelectric materials lately and I've steel looking for an answer for a major question -
Is there any way to calculate the watt output of a certain piezoelectric component* ?


*The piezoelectric components I'm talking about are those which convert mechnical force to electricite ( for example , tiny piezoelectric components which make a led turn on once they're touched , like this one : http://www.youtube.com/watch?v=Xuw9frP1GNo

Thanks in advance
 
  • #17
411
2
Earth ground believed to be good electrical yet poor thermal conductor.
Is it possible to use in some thermoelectric cycle?
For example we could make some large heat resistant artificial objects to swim in
magma underground and establish some electric resonance between them and
some devices on Earth surface?Could energy be transfered between hot and cold body
with help of AC current?I dont know but there is some telluric current registered in the
ground.The is some thermoelectric currents believed to be in Earth too.
 
  • #18
I need dipole moment of anisotropic dielectric sphere
The dielectric tensor is given in cartesian matrix
I try to find the potential inside and outside by expanding spherical harmonics
But sinece the dielectric tensor is cartesian
I have a cartesian derivatives of potential in electric induction D
When I plug it in the boundary condition
D normal is continuous
I end up with nonlinear spherical harmonics (inverse or squared or cubed)
So, I can not find the coefficients and stuck

Please help
It very urgent
 
  • #19
Hello,
I want to derive the electric field vector from a quadrupole potential.
I am starting from
faculty.physics.tamu.edu/pope/em603.pdf (6.32)
and I am considering E=-∇[itex]\phi[/itex]
with [itex]\phi[/itex]=1/2 Qijxixj/r5
I am very unsure if I carry out the thing properly, I would appreciate your comments:

Ek=1/2 ∂k(Qij xixj/r5)
=Qijkxixj/r5 - 5xkxixj/r7
=Qijkinj-5nkninj)/r4

with ni=xi/r

Now, what is the meaning of δkinj ?! Can it be anyhow simplified?

Thanks for reading!
 
  • #20
1
0
I have a question that involves two electric dipoles that are situated at the origin and are perpendicular to each other. Other than that they are identical in every way. Each one oscillates so at cos (w(t-r/c)). I'm asked which multipole moments oscillate. Wouldn't it just be the dipole moment and nothing else. Or would a quadrupole moment work. I can find the E and B fields through superposition and the radiated power, but all of my fields are still 1/r dependent. What am I doing wrong?
 
  • #21
1
0
Hey everyone,
I came across this recently through a friend. I understand how a regular homopolar motor works, but I simply can't figure out how you get the copper wire sitting on top of the cathode of the battery to spin without a connection to the anode!

I made a short video, please search "VsBePtKUyyw" on youtube, I can't post links yet.

My main question is:
Where and how does the current flow in order for the copper wire to experience a Lorentz force and start turning?

When the magnet is attached to the anode or "- terminal" of the battery, the battery itself becomes a magnet. By placing a certain shaped copper wire on the cathode without completing the connecting to the anode, the wire begins to spin. I've looked everywhere and have not been able to find an example like this, all the other homopolar motors have a completed or loop wire connection from cathode to anode.
Can anybody explain this without pasting links in here??
 
  • #22
Hi! This is a quite sophisticated problem, but it’s fun and interesting!

A coaxial current transmission line is short-circuited by a cylindrical disk in one of its ends. Let’s say that the disk and the coax cable have the radius [itex]r_{2}[/itex] and that the disk has the thickness [itex] d[/itex]. We’re interested in solving the (complex) electric field [itex]E_{z} [/itex] directed in the [itex]\hat{z}[/itex] direction in this disk. When you work Maxwell’s eq. out you’re arriving at the PDE:

[itex]\nabla^2 E_{z} - j\omega\mu\sigma E_{z}=0[/itex]

where [itex]\sigma [/itex] is the conductivity, [itex]\omega[/itex] is the frequency, [itex]\mu[/itex] is the permeability and [itex]j[/itex] is the imaginary unit. This disk has cylindrical rotation symmetry so [itex]E_{z}[/itex] does not depend on [itex]\phi[/itex]. If we choose our cylindrical coordinate system so that the z-axis passes through the center and that z=0 at the circular plane of the disk not in direct contact with the coax cable; then the boundary values on the disk are the following:


1. [itex]E_{z}(\rho,z=0)=0[/itex] for all [itex]\rho\in[0,r_{2}][/itex]. (because there is a current on the circular plane surface not in contact with the coaxial cable).
2. [itex]\int_0^{r_{2}}E_{z}(\rho,z’ )\rho\,d\rho=0[/itex] for all [itex]z’ \in(0,d)[/itex]. (because the total current through a cross-section of the disk is zero).
3. [itex]\int_{r_{1}}^{r_{2}}E_{z}(\rho,z=d)\rho\,d\rho=-I/2\pi\sigma[/itex], where [itex]0 \leq r_{1} \leq r_{2}[/itex] the inner radius of the shield of the coax and [itex]I[/itex] is the given complex current. (because the current through the shield of the coax must be –I)
4. [itex]E_{z}(r’ ,z=d)=0[/itex] for all [itex]r’ \in[r_{0},r_{1}][/itex], where [itex]r_{0}[/itex] is the radius of the inner leader, such that [itex]0 \leq r_{0} \leq r_{1} \leq r_{2} [/itex]. (because we have a current on this surface)
5. [itex]\int_0^{r_{0}}E_{z}(\rho,z=d) \rho d \rho=I/2\pi\sigma[/itex]. (the total current penetrating into the disk from the coax cable is of course I.
6. Obviously [itex]E_z[/itex] must also be finite for all points in the disk.

My problem is to solve this PDE with these boundary values. I would be very thankful for any insight or idea on how to solve this problem (full solutions not necessary acquired!). So if you can help me in any way I owe you a huge amount of thankfulness and respect!
 
  • #23
11
0
From an EM textbook (dealing with an above surface time varying magnetic field source):

Primary field: Field artificially produced.
Secondary field: Field produces by the charge/current created by the primary field.

“Because air is so highly insulating in comparison to almost all in-situ earth materials, the conductivity contrast at the earth’s surface is virtually infinite… Since the normal component of total current density (conduction + displacement) must be continuous through a boundary, a charge distribution and a corresponding jump in the normal component of the electric field must occur there. The secondary electric field of the interface surface charge neutralizes most of the electric field that would otherwise exist within the earth, and it increases the electric field in the air. The only part of the electric field which continuous to be found in the earth is a circulation about the primary magnetic field lines which penetrate into the earth. A corollary of this reasoning is that an above-surface time varying magnetic source cannot produce an appreciable component of the electric field or current in a stratified earth normal to the layering”

The bolded part is the part that I do not understand.

My interpretation of the first part:

Layer 1 (Dielectric, air, conductivity = s1 = 0) : Layer 2 (conductivity = s2)
Jn1=Jn2 Current density continuation equation (n is the normal component)
E1n*s1=E2n*s2 Ohms law
0=E2n*s2 Since Jn1=0
E2n=0 Since s2 =/= 0

Thus an above surface time varying magnetic field source cannot produce an electric field normal to the boundary. Thus the electric field within this stratified earth will be purely tangential to the boundary.

However, then it says that “the only part of the electric field which continuous to be found in the earth is a circulation about the primary magnetic field lines which penetrate into the earth”. I interpret this as Curl E = -dB/dt. But isn’t that the electric field were dealing with in the first place? The above ground time varying magnetic source produces the electric field via Curl E = -dB/dt. And surely it is not hard to imagine that Curl E = -dB/dt will produce an electric field that isn’t tangential to the boundary everywhere? Can anyone shed some physical meaning onto this for me? It seems as if they are talking about more than one electric field, one which circulates about the magnetic field lines (their time derivative) and then another which is having its normal component nullified.
 
  • #24
4,742
102
I have a few question in the Image Theory with electric and magnetic conductor plane. The drawing below is from Advanced EM theory by Balanis p 318. The top drawing are vertical and horizontal electric and magnetic sources on top of an electric conductor plane where σ=∞. The bottom drawing are vertical and horizontal electric and magnetic sources on top of a magnetic conductor plane.

[PLAIN]http://i39.tinypic.com/2r7wumx.jpg[/PLAIN]

My questions are:

1) Using the case with electric conductor plane as the top drawing. Say the plane is copper that is electrically conductive by not a magnetic conductor. I fully agree with the drawing of the electric sources. But if the plane is not even magnetic conductor, why is there even image for the magnetic sources? Is this because it is assume time varying and the image magnetic source is accompanying the image electric source ( as they cannot be separated)?

2) A lot of magnetic conducting materials are also electric conducting material( eg. Carbon steel with high μ). So if you have that as the plane, you have a situation that the plane is both electric and magnetic conducting. How are the images going to look like? Confusion is because the direction is opposite in both case and they cannot be summed together. They cancel!!!

Thanks
 
  • #25
Hello,

I would like to know what exactly the concept is behind electric/magnetic moments.

Just to give a little context, I'm currently doing an internship for the summer and my research deals with toroidal dipoles in metamaterials. I'm just having a hard time understanding what a toroidal dipole does to light that is different to electric/magnetic dipoles.

I know electric/magnetic moments can be used to calculate the fields and energy of dipoles, but I'm having a hard time understanding what a toroidal moment can be used for. Calculating the field or energy of a toroidal dipole? I figured that if understand the main concept behind electric/magnetic moments, then maybe I could better understand the toroidal moment. From what I've read, toroidal dipoles are non-radiating, so does that mean they do not generate fields around them?

I've read that metamaterials that generate a toroidal dipole have low energy loss. If anyone is knowledgeable of metameterials and how they work, could you also explain why that is and what that means? Like what's happening to the electro-magnetic wave as it passes through a metamaterial that is structured in a way that generates a toroidal dipole and how is that different to metameterials that only generate magnetic dipoles?

Thank you very much in advance for taking the time to help me.
 

Related Threads on Electric field boundary equation implication at air/earth interface

Replies
0
Views
1K
Replies
1
Views
672
Replies
2
Views
6K
Replies
11
Views
1K
Replies
7
Views
6K
Replies
3
Views
4K
Replies
6
Views
2K
Replies
1
Views
568
Top