# Pleeeease Help Why a dielectric absorbs only a certain amt of field

1. Jun 6, 2007

### Mr Virtual

Pleeeease Help!! Why a dielectric absorbs only a certain amt of field

Hello Everyone!

I am a new member in this forum.

What we learn at school regarding dielectrics is: When you apply an electric field on a dielectric, the molecules of the dielectrics polarize and stretch (the electrons/anions shift towards the positive terminal, and the cations shift towards the negative terminal), which creates an opposing field, which reduces the net applied electric field.

This is what I think the above statement means : that if you have two metal plates separated by a dielectric medium, and apply an electric field across the two plates, thenthe field travels from the positive plate, enters the dielectric, does work in polarizing/stretching its molecules, gets reduced and reaches the opposite plate with this reduced value.

But I ask: why does the electric field only do a limited amount of work? Why doesn't the field stretch the dielectric's molecules to the fullest extent possible for it? Since electric field is also a kind of force, why isn't this whole force converted to work done in stretching the dielectric.

According to me, the molecules of the dielectric should consume the whole applied electric field and stretch accordingly and, thus, no electric field should ever reach the negative plate, as all the field is converted to work done in stretching. This is what happens in an insulator.

Then what makes the dielectric absorb a limited amount of field and let the rest of it pass through.

Please answer me this question in any way you know (quantum approach, QED or classical approach), but make it with minimum or no mathematics, because I want to understand the theoretical portion, and also because since I am still school student, I won't understand the complicated quantum maths much.

Love to you all and warm regards.
Mr Virtual

2. Jun 7, 2007

### Mr Virtual

I got it

Hello again

Is my question boring or something? If it is, I am sorry I put it up.

Never mind. I think I have found the answer to my question. And it is quite simple, the way I look at it.

Dielectrics, like insulators, absorb all the electric field. What differentiates them from insulators is their ability to stretch in a given external field.

Suppose we have a single positively charged metal plate of infinite length, exerting a field x in both directions:

3. Jun 9, 2007

### Fizik

hi, in principle an ideal insulator will never depolarize any electric field because by definition the charges are not allowed to move around their initial position. You can think of this ideal-ness as supplied by other sources of restoring force such as nuclear strong force of just simply an overwhelming strong electrostatic force.

On the other hand, an ideal conductor will completely depolarize the electric field because by definition the conductor can supply free charge (delocalized electron sea) to do it.

4. Jun 10, 2007

### damgo

Hi MV,

You got it, though I wouldn't talk about dielectrics etc. "absorbing" electric fields. When we apply an electric field, we put that field in and that field is there -- that's what we mean. It doesn't get "absorbed" or anything like that. Rather, this field causes the dielectric material to polarize. The polarized molecules create their own electric field, which happens to be opposite to the applied field. So the total field you get is smaller than the applied one.

But you're absolutely right that what makes the difference is how much the charges in the material can "stretch." In a conductor, charges are basically completely free: they will move around to completely cancel out an imposed electric field. (This is why the E field is always zero inside a conductor!) In a dielectric, molecules can stretch a little, but they are still attached together. Think of them as being two little charges (one + and one -) attached by a spring or rubber band. This attachment force lets the molecules stretch to cancel out a little of the applied field, but stops them from moving enough to completely cancel it.

5. Jun 11, 2007

### Mr Virtual

I think electric field is absorbed by dielectric

Dear Fizik and Damgo, thank you so much for your replies. I was losing hope.

I agree with damgo's explanation because it is the same as what we are taught at school viz. E = Eo - Ep

But mate, you will agree that stretching of a dielectric is directly proportional to applied field i.e. more the applied field, more is the stretching. Also, since the applied field causes stretching, we can say that work is done by the field or, in other words, energy is spent by the field. As intensity of field is increased, more energy is provided by it to the dielectric, resulting in greater stretching. When I said "absorption" of field, I actually meant absorption of this energy by the dielecetric, causing stretching.

So, now we can state your explanation in another way: When a field is applied to a dielectric, some of the field's energy is consumed in stretching ( that is, in cancelling the internal opposing field of the dielectric), and the rest of it is preserved in the field.

Let us now apply this concept. Suppose we have a non-polar molecule like that of nitrogen ( or any other diatomic molecule ). When field is applied to this molecule, stretching of this molecule is actually the distancing of the valence electrons ( in the [sigma (2pz)]2 orbital), from the remaining kernel(nitrogen nucleii+ electrons other than valence electrons).

Now suppose we apply a field of 2 N/C (all of these values are imaginary). The energy supplied by this field is, suppose, 20 J. The nitrogen molecule should absorb all the 20 J and stretch. But according to your explanation, the nitrogen molecule absorbs only a part of the applied 20 J, let's say it absorbs only half of it i.e. 10 J of energy or 1 N/C of field. Due to this, it stretches only half of how much it would have stretched if it had absorbed 20 J.
Now I ask you, why doesn't the molecule absorb the whole of the 20 J of energy? What is preventing it to do so? Or, in other words, why doesn't the molecule absorb the whole of the field? ( as I asked in my original post). As I have said already, stretching in nitrogen is caused due to shifting of valence electrons and kernel away from each other. As far as I know, both electrons and protons absorb all the energy that is provided to them, whether it is due to electric field or heat. Then why are electrons and protons of a dielectric so special that they only partially absorb the applied field? This is what confused me so much in the first place. So I thought of trying to find out a different explanation.

According to me, when electric field of 2 N/C is applied to the nitrogen molecule, it absorbs whole of the 20 J of energy, and stretches upto the point where internal field ( field between kernel and valence electrons) becomes equal to 2 N/C i.e. the internal and external field balance or cancel each other. The net field of the molecule, it appears, should be zero. But it isn't. When the molecule is stretched, distance between the kernel and the valence electrons increases, which means the kernel becomes slightly positively charged and the electrons become negatively charged (obviously). The fields due to these two poles of the polarized/stretched molecule add to give a net field in the direction of the applied field. This is the field that reaches the opposite plate.

applied field
--------------------->
- stretched molecule +
<---------------------
internal field

Applied field is cancelled by the internal field.
What remains is the stretched molecule.

stretching causes the two poles to develop their own small fields

----> - stretched molecule + --->

net field
---------->

Net field is in the direction of applied field.
This field reaches the other side of the dielectric.

This field is always less than the applied field. More the stretching, more is the polarity of the molecule, and stronger is this field. That is why non-polar molecules like nitrogen, which have weaker bonds, stretch greatly and thus a strong field ( almost equal to the applied field ) reaches the other side of the dielectric. Whereas, in polar molecules like water, in which bond is strong, stretching is less, and hence, weaker field is created.

So, we now know that a dielectric does not absorb the applied field partially, it absorbs it fully. Due to stretching, poles are created which then create a field that finally reaches the other side of the dielectric.

An insulator does not stretch, so no poles are created and thus, no field reaches the other side of the insulator.

Please tell me where I am wrong.

Thanks a lot!!

Warm regards
Mr Virtual

6. Jun 11, 2007

### damgo

^^^ You're right in that when we impose a field on a dielectric, work is done on the dielectric itself. But we need to be careful about terminology here. When we talk about imposing a field, that means we set up charges in order to create the specified field (or in this case, what would be the specified field if no material were present.) Because of the energy going into the dielectric, the amount of work that needs to be done to set up this field changes... but the imposed field itself isn't 'absorbed' in any sense. In this situation, we have a fixed imposed field, which can require any amount of energy to set up. We don't have a fixed amount of energy which can be "used up."

Onto the central point, the mistake is that objects don't "absorb all the energy that is provided to them." I'm not sure where this came from; it's certainly not true. (It's not even meaningful -- what would it mean? Say we couple two systems together; which one will absorb all the energy? They can't both do so!)

If you want to think about the microphysics of what's going on, consider a simplistic model of a dielectric molecule. We can distort the atom in order to polarize it. But in doing so we're pulling the molecule out of its lowest-energy configuration, and hence there's a restoring force that resists us. At some point this force will become sufficiently strong to stop the molecule from polarizing further.

It might help to think of a toy system of two charges attached by a spring. If I place this system in an electric field, the degree to which it polarizes depends on the spring constant ( dx ~ E_eff*q/k). So in the limit that k is very large, we get no polarization; if k is small, the polarization will be nearly complete.

7. Jun 11, 2007

### Mr Virtual

Your example of the spring toy was heplful.

"We can distort the atom in order to polarize it. But in doing so we're pulling the molecule out of its lowest-energy configuration, and hence there's a restoring force that resists us. At some point this force will become sufficiently strong to stop the molecule from polarizing further."

Hey! You have yourself stated that the spring will stop stretching further after a certain point, and this is going to happen only if the applied force is balanced by the restoring force. If you go deeper, we can say that the atoms of the spring "absorbed" the energy provided by the applied force, and kept stretching untill the restoring force became high enough to stop further stretching. You cannot say that if we are applying 20 N force on the spring, it will absorb energy equivalent to 15 N force and stretch no further, even if it can easily stretch on 20 N force. It is not right to say that it will require 25 N force to stretch equivalent to 20 N. The spring is going to absorb the whole energy equivalent to 20 N. That is what I meant when I said that a system will absorb all the energy provided to it.

According to me, whenever work is done on a particle or a system, the energy is stored or "absorbed" in it.
In my previous post, I am saying "absorption of field" because I usually consider the field in the quantum electrodynamical sense i.e. an electric field is made of invisble quanta of energy called virtual photons, which an electron or any charged particle definitely absorbs and responds by moving away from or towards the direction from where the photon came.

Let us consider the particular case of a parallel plate capacitor, with no air as dielectric.
When a field Eo is set up between the two plates, it means that the photons of a particular energy are now getting exchanged between them. Since field is uniform, the density of photons is same everywhere between the plates.

Now when we introduce a dielectric molecule in this capacitor, the molecule absorbs all the photons coming its way and responds by polarizing/stretching.
Since these photons contain energy, this energy is gained by the electrons and the kernel, and they come out of their lowest energy state. The photons from the positive plate cause repulsive action on the kernel and attractive action on electrons, and the opposite is true for the negative plate.

So,
1. Field is absorbed by a dielectric molecule
2. The field provides definite energy to the molecule, since it is uniform in nature. The energy depends on the strength of the field.
3. This energy stretches the molecule, untill it balances the internal field of the molecule.
4. In this way, whatever field the molecule is receiving, it is getting "absorbed" ( if you take the 'virtual photons' approach)
5. Distance between the kernel and valence electrons increases, and thus attraction decreases. We can suppose that a molecule of nitrogen has become somewhat like a water molecule at this point i.e. it has developed (weak) poles. If you add the field due to these poles, we get a net field in the direction of the applied field. (Remember, the internal field of the molecule has already been cancelled by the applied field. So there is no field opposing this field developed due to polarization of the molecule).
6. This is the field which reaches the opposite plate, but we assume it to be a part of the partially-cancelled applied field.

Please forgive me for being repetitive, but I wanted to make the clearest impression of what I am saying.

Mr Virtual

8. Jun 12, 2007

### damgo

Don't confuse energy with force! They're different concepts. Let's draw out our toy model a bit more. If each end of the molecule has charge q, and the imposed field is E (ignoring other molecules for the moment), then the work done on the molecule (neglecting the electric attraction of the molecule itself) is

Work = 1/2 * k*dx^2 ~ E^2*q^2 / k

Note that this depends on k, which has nothing to do with the imposed field. Indeed, in the limit as k -> infinity, the energy in the molecule goes to zero. It's only in the limit that k->0 that the "absorbed" energy of the molecule will approach that of the local imposed E field.

---

While it's possible to view static fields as being composed of photons, it's not straightforward and often not helpful. For example, no static field configuration can possess a definite number of photons. Indeed IIRC for electrostatic fields we have to include states with unbounded (ie infinite) numbers of photons present. In this case, where no quantum effects need to be introduced to understand the basic phenomenon, trying to understand static fields from a virtual photon standpoint is particularly unhelpful... Naively thinking of "absorbing" and "emitting" photons is going to lead your intuition in exactly the wrong direction. (For example, opposite charges attract, but it is not easy to see how this comes about through the emissions and absorption of virtual photons!)

9. Jun 12, 2007

### Mr Virtual

A great thing about this forum is that it has got people like you who are always ready to clear doubts of people like me.
Thanks a lot Damgo, and Fizik, for your help. I have now completely understood what you (Damgo) were trying to explain to me for the last one week.

Thanks again
Mr V

10. Jun 13, 2007

### Mr Virtual

Damgo
If you are still online, please don't leave. I have got a few doubts to clear about your spring-toy explanation. I will start in a few minutes.

regards
Mr V

11. Jun 13, 2007

### Mr Virtual

My question

I will post my question in a short while.

Last edited: Jun 13, 2007
12. Jun 13, 2007

### Mr Virtual

I am rewriting my question in my next post.

Last edited: Jun 13, 2007
13. Jun 13, 2007

### Mr Virtual

I now understand how 'springy' some molecules can be. But this has got me stuck into the working of a simple spring itself.
If we stretch a spring, connected to a wall, by x metres then
F = -kx
W = 1/2*kx^2

When F newtons of force is applied on this spring, it stretches until restoring force becomes equal to -F, thus exactly cancelling the applied force. The amount of stretching depends on k.
As k keeps getting larger, x becomes smaller. But x can never be zero until k is infinite (which is not possible). Since displacement is always there on application of force, energy is always spent if force is applied. And this energy stored in a spring is always the same (at a given force), only the amount of displacement x differs with change in k.

Similarly, in molecules, if a field E is applied, then if k is higher, stretching is small and if k is smaller, stretching is big.
But the amount of energy stored must be the same. If I am wrong, please clarify where.

A question: if molecules are springy, then when we impose a field E, why do they also not stretch until their restoring force equals -E, thus cancelling the applied field. What I mean is, why do they cancel the applied field only partially? What stops them from stretching to their maximum capacity, like a normal spring? Again, please clarify.

If your formula for energy stored in a spring with charged bodies as end points is totally different from the normal equation of spring, then I am sorry, I haven't yet studied this formula and as such I can't make out what '~' stands for in it.

regards
Mr V

Last edited: Jun 13, 2007
14. Jun 13, 2007

### damgo

What makes you think the energy stored in a spring (for a given force) is constant? If we write the energy stored as a function of F and k, we get

x = F/k
W = 1/2*kx^2 = 1/2*F^2/k.

Intuitively, it might help to remember that energy is the integral of force (w.r.t. displacement.) Under the application of some constant force, how far a system responds is determined by how quickly the restoring force increases with displacement. But the energy stored is the integral of that restoring force wrt displacement; so in general a slower-rising force will lead to more stored energy.

The symbol ~ generally means "approximately equal to" or "proportional to." I used it instead of the equals sign to indicate that I wasn't keeping track of factors of 1/2 and so on.

15. Jun 14, 2007

### Mr Virtual

Damgo
I need to clarify a few more things about springs, so I am creating a new thread (this thread was actually related with functioning of dielectrics, not springs). After that, I will come back to our discussion.

regards
Mr V

16. Jun 28, 2007

### Mr Virtual

Sorry for the long delay.

So, a dielectric works as a spring. It decreases the field between the two plates of the capacitor. Lesser the value of k, more will be the decrease in field (k is the Hooke's constant, if we consider the molecules of dielectric to be acting as tiny springs).

Now k is decided by the bond strength of the molecules of the dielectric. Right? Greater the bond strength, greater is the value of k, and lesser is the decrease in applied field. This means molecules which are "more easily stretchable" (having lower k) have higher dielectric constant than "less easily stretchable" ones (having higher k).

1. strong bond = high k = less polarization/stretching (per unit force/field applied) = small decrease in applied field = low dielectric constant

2. weak bond = less k = more polarization/stretching = considerable decrease in applied field = high dielectric constant

Polar dielectrics can be considered similar to already-stretched springs, so stretching them more by applying a field results in further stretching. As already told, more the stretching more is the decrease in applied field. So polar dielectrics reduce applied field drastically.

Am I right so far?

regards
Mr V

17. Jun 28, 2007

### Mr Virtual

Since nobody has answered yet, I will continue.

As already told by me, *stretching* is the actual cause of decrease in applied field. No stretching means no decrease. This means that molecules which do not stretch or stretch negligibly under an electric field, should actually transfer the whole field from one end to the other (like if we lift an object with a rod, which (of course) doesn't stretch, the whole force applied by us is transferred to the object).
Now, insulators do not stretch in a given electric field. Then they should let the whole field pass through. But this is not so; they do not allow any field to pass through, which impiles they should stretch (but they don't).
Am I wrong in suggesting that insulators don't stretch? Is reality the opposit of this: that they stretch the most, and therefore, they do not allow applied field to pass through?

Mr V

18. Jun 28, 2007

### Claude Bile

Insulators ARE dielectrics. Think about it, dielectrics are media that posses no free charges. Insulators, too are defined as possessing no free charges (which is why they conduct electricity and heat so poorly compared to conductors).

Claude.

19. Jun 28, 2007

### Mr Virtual

No, not if you go deeper. Dielectrics are insulators, but insulators are not dielectrics. Dielectrics are insulators which allow some of the applied field to pass through them. Insulators do not allow any field to pass through. So, you see - insulators are not dielectric. It is like dielectrics are special type of insulators.
But my question was: whether insulators stretch to a great extent (thus, never allowing any field to pass through), or whether they do not stretch at all?

regards
Mr V

20. Jul 1, 2007

### Claude Bile

Wherever did you hear such a thing? Can you provide a reference? The best interpretation of this statement I could make is that you think there are substances where the entire applied E field is reduced to 0 inside that substance. Such substances do exist, they are called conductors. Are you saying conductor = insulator .

Not to mention the implicit contradiction in the statement that "dielectrics are special types of insulators" despite the fact their definitions as you have presented it are mutually exclusive.

Claude.

21. Jul 2, 2007

### Mr Virtual

Yeah, I think I said the wrong thing.
So,
1. Insulators polarize on application of applied field, thus creating an induced field, which partially cancells the applied field.
2. If induced field is small, then net field (applied field - induced field) is considerably high, and such an insulator can be used in capacitors. Such insulators are also called dielectrics.
3. If induced field is high, then net field is small.
4. An ideal insulator will undergo extremely high polarization on applying external field. Thus, almost no net field will reach the other side of insulator.
The field inside the insulator will be close to zero (since, induced field will almost cancel the applied field), but not exactly zero.
5. In case of a conductor, induced field will exactly cancel the applied field. However, unlike insulators (in which electrons are bound), conductors have free electrons which reach the other side of the conductor in order to polarize. These electrons create both an internal field (which cancels the applied field), and an external field reaching the other plate of the capacitor.
In dielectrics, only the internal field is created. Thus, although conductors completely cancel the applied field, they still seem to allow all the applied field to pass through it, when in reality it is the field created by the displaced electrons.

Comparing with springs:
1. Dielectrics (non-polar) have high k (Hooke's constant). Thus, their stretching/polarization is quite limited in a given external field.
2. Insulators (non-polar) have low k. Thus their stretching/polarization is quite high in the same external field.
3. Polar dielectrics/insulators have natural polarization, which automatically cancels a large part of the applied field. The rest of the applied field is cancelled as is explained in 1 and 2.
4. For conductors, k~0. Thus, their polarization is almost infinite.

Am I right?

regards
Mr V

22. Jul 2, 2007

### Claude Bile

1. Agreed, though the word "opposes" is more correct than "cancels" because cancels carries with it the implication that there is nothing left.
2. Materials with a higher permittivity and thus a higher induced field are better intermediate layers for capacitors in terms of storing charge because the higher induced field results in a lower net field, which allows the capacitor to pack more charge on its plates.
3. Yes.
4. No, an ideal insulator will undergo zero polarisation because in a perfect insulator, the charges are completely immobile. This would result in zero dampening of the applied field. The situation as you describe it is physically impossible, since it implies we can cram an infinite amount of charge on the plates of a capacitor.
5. Loosely correct, except in reference to the insulator.

I think the main point to be made is number 4, a perfect insulator has zero charge mobility, in other words the charges in a perfect insulator are completely immobile, not, as you seem to think, infinitely mobile yet bound.

Claude.

23. Jul 3, 2007

### Mr Virtual

Yeah, that is the main point that is troubling me. Actually, damgo first suggested me to compare the working of a dielectric with the working of a spring.

Now,
1. if spring's k is low, stretching is high and applied force decreases drastically. Similarly, if polarization is high, then external field falls sharply.
2. if spring's k is high, stretching is low and applied force decreases only by a small measure. Similarly, if polarization is low, then applied field decreases only slightly.

But if you consider a solid rod, it does not stretch at all, because it has almost infinite k. Thus, applied field does not decrease at all. Here I am stuck up. If you apply this analogy to dielectrics, then an insulator does not stretch at all too, which should mean that external field is not reduced, which means that almost all the field should reach the other end of the insulator. But this does not happen. Why? Am I wrong, or is the spring-analogy not applicable in this case? How would you exactly describe the absence of electric field without the absorption of any?

धन्यवाद (thank you)

regards
Mr V

Last edited: Jul 3, 2007
24. Jul 3, 2007

### Claude Bile

Why do you insist that the electric field must be completely absent inside an insulator? Maybe you are confusing the fact that a perfect insulator generates a negligible field in response to the applied field, and not that the applied field itself becomes negligible.

Consider that a vacuum is a very good insulator, and that one would not expect any reduction of an applied E-field at all due to the absence of any dipoles.

You're welcome .

Claude.

25. Jul 3, 2007

### Mr Virtual

Actually I am in a hurry, I have to get ready for school. I will answer you when I come back.

देरी के लिये क्षमा करे !(Sorry for the delay)

Mr V