Understanding Dielectric Polarisation: Transformation of Dipoles and Charges

[gracy]

A dielectric is placed between two capacitors ,How (a) gets transformed into (b)

Where did all the dipoles present in picture (a)go?And from where the negative and positive charges appeared inside the dielectric in picture (b)

 

[cnh1995]

The + and - poles in the middle region will cancel each other. That will leave +ve polarity near -vely charged plate and -ve polarity near the +vely charged plate. Its a rearrangement of the charges inside the dielectric. I believe its called dipole polarization.

 

[gracy]

The + and - poles in the middle region will cancel each other.

Then these dipoles should remain which I have shown in blue box.

 

[cnh1995]

Then these dipoles should remain which I have shown in blue box.
View attachment 93002

Their poles near the walls cause the electric field inside the dielectric.

 

[gracy]

I still don't understand.

 

[cnh1995]

I still don't understand.

Consider left blue column. Its +ve poles will nullify the effect of the -ve poles of the next column. This will go on till the last column(rightmost blue). The +ve poles of the last column and -ve poles of the first (leftmost blue) column will create the electric field inside the dielectric which will be opposite to the one created by the plates.

 

[gracy]

Got it!Thanks

 

[cnh1995]

Got it!Thanks

Well that's the overall picture of how it works. Its not in proper physics language so I guess gneill or some other expert will later come up with a way better explanation. Until then, hope this is fine.

 

[cnh1995]

Consider left blue column. Its +ve poles will nullify the effect of the -ve poles of the next column. This will go on till the last column(rightmost blue). The +ve poles of the last column and -ve poles of the first (leftmost blue) column will create the electric field inside the dielectric which will be opposite to the one created by the plates.

I'm not sure if "nullify" is the proper word here. I believe it should be "attract".This structure forms a resultant single large dipole that opposes the field due to the plates. Think of the dipoles as batteries connected in series. Their voltages will add and across the two far ends, this total voltage will be available. Similarly, the fields of the dipoles will add and the resultant field will be available inside the dielectric ,opposite to that of the plates.

 

[Dale]

Where did all the dipoles present in picture (a)go?And from where the negative and positive charges appeared inside the dielectric in picture (b)

It is just showing that the dielectric material has no bulk volumetric charge density, but does have a surface charge density.

 

[gracy]

I know The electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field.I t is denoted by χe But I want to know what does this mean
The electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field.

My question is what is ease / difficulty when it comes to polarization of dielectric?

 

[sophiecentaur]

My question is what is ease / difficulty when it comes to polarization of dielectric?

The effect of introducing a dielectric - with its easily polarised molecules is to increase the Capacitance between two conductors (plates). That means the charge difference between the plates will be much greater for a given PD across them. The charges on the faces of the dielectric - and the charges on the plates - correspond to what you would get if the plates were much closer together in air. The practical advantage is that a thick layer of insulating dielectric will allow a much greater operating voltage (hence a greater stored charge for a given capacitance) and reduce the requirement for the plates to be accurately flat (i.e. the percentage accuracy of the spacing for the area). The 'ease' is proportional to ε_r (>1) because, of course, the hardest is a vacuum (ε_r = 1). The easiest is a metal conductor. Unfortunately, a conductor will polarise with no PD across it and allow the charge to pass through it so you would also need an air gap in series, to use it in a capacitor.

 

[cnh1995]

My question is what is ease / difficulty when it comes to polarization of dielectric?

Suppose the field established by the charges on the plates is Eo and there is no dielectric, just air(or vacuum). Now if a dielectric slab is inserted, it will be polarized and create its own opposing field E1 which will reduce the net field strength in the dielectric.
So, net field inside the dielectric is now,
E_net=Eo-E1
and the ratio Eo/E_net is called as the relative permittivity ε_r or the dielectric constant.
From this you can see why ε_r=1 for vacuum and ∞ for conductors.
With increase in ε_r, degree of polarization increases and net field between the plates decreases. That's what the word "ease" describes here.

 

[cnh1995]

electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field

ε_r=χ_e+1.
So, as the susceptibility increases, relative permittivity increases, easing the polarization.
You can see, χ_e=0 for vacuum since it is the least(or not at all) susceptible dielectric.

 

[just dani ok]

I know The electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field.I t is denoted by χe But I want to know what does this mean
The electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field.

My question is what is ease / difficulty when it comes to polarization of dielectric?

Something is called easy when a result can be achieved with less effort. In this case the result is the polarization of the material which in turn determine total electric charge stored in a capacitor while the effort is the potensial required to polarize the material for required quantity of electric charge.

 

[nasu]

It may help if you look at the actual formula defining the susceptibility.
$\vec{P} = \chi \vec{E}$
A large susceptibility means that you can get the same polarization with a lower field. So it is "easier" to get it, isn't it?
There are many similar cases.
Mass measures "how easy" is to accelerate something (F=ma).

Note: In SI the formula above contains the permitivity $\epsilon_o$ of free space too but this does not change the relationship.

 

[gracy]

A large susceptibility means that you can get the same polarization with a lower field. So it is "easier" to get it, isn't it?

And how it (susceptibility)does that?I know it is property of a material but how susceptibility contributes in polarization of a material?

 

[sophiecentaur]

And how it (susceptibility)does that?I know it is property of a material but how susceptibility contributes in polarization of a material?

"How"? Do you mean the mechanics of the material?

 

[cnh1995]

And how it (susceptibility)does that?I know it is property of a material but how susceptibility contributes in polarization of a material?

I believe that depends on the "chemistry" of the material. There are various types of polarization e.g. dipolar , electronic, ionic polarization etc.
In dipolar polarization(the one shown in #1) ,dipoles are created in the dielectric in response to the applied electric field ,resulting in an opposing field, thus reducing the original field.
P=χ*E gives the relationship between polarization, susceptibility and the net electric field(E) in the dielectric.
If no opposing field is generated (e.g.vacuum), P=0(no polarization), thus giving χ=0 ( i.e. vacuum is not susceptible to the applied field) and in case of conductors, E=0 (i.e. they are too much susceptible!), thus giving χ=∞. Other dielectrics show intermediate susceptibility.

And how it (susceptibility)does that?

I would say susceptibility doesn't do that but the materials do, by means of their chemistry and applied electric field. This behavior of the materials to respond to the applied electric field is quantified by "defining" what we call as their "susceptibility". It is a constant "determined" as the constant of proportionality in the experimentally obtained equation, P∝E_net.

 

[Dale]

And how it (susceptibility)does that?I know it is property of a material but how susceptibility contributes in polarization of a material?

I don't think that this question is what you want to ask. Susceptibility is simply defined this way, so the question is a little like asking how mass makes a particle more difficult to accelerate.

I think what you probably actually want to know is what are the microscopic properties in a material that make it have high susceptibility. This question is more like asking what about a material makes it have high mass.

 

[gracy]

What is difference between permittivity and susceptibility?I find them same.Because dipole creation, electric field reduction, and surface charge density are all ways to interpret the effects of permittivy. A permittivity value just tells us how strong the effects are.And that's what susceptibility indicates!

 

[gracy]

vector $E$=vector $E_o$-vector $E_p$
vector $E$=$\frac{E_o}{K}$
Is the following relation correct
K($E_o$ -vector $E_p$)=$E_o$

 

[cnh1995]

A permittivity value just tells us how strong the effects are.And that's what susceptibility indicates!

Right. But permittivity gives the relation between net field and the original field while susceptibility gives the relation between polarization and the net field. So, according to my understanding, permittivity tells you how "strong"(may not be the correct word) a dielectric is w.r.t vacuum( free space) while susceptibility tells you how sensitive a material is to the applied electric field. Susceptibility is a general term and doesn't depend on vacuum. But permittivity is first defined for vacuum and then applied to all the other materials (using relative permittivity or the dielectric constant).
Susceptibility measures the sensitivity of the dielectric to the applied field through polarization while permittivity tells you how much charge is required(or permitted) to create the same electric field in different dielectrics.

 

[cnh1995]

vector $E$=vector $E_o$-vector $E_p$
vector $E$=$\frac{E_o}{K}$
Is the following relation correct
K($E_o$ -vector $E_p$)=$E_o$

Yes.

 

[gracy]

If I will give charge density σ to a dielectric and then will give same charge density to different material,will vector $E$ be different in each of them?

 

[cnh1995]

If I will give charge density σ to a dielectric and then will give same charge density to different material,will vector $E$ be different in each of them?

Yes. That is what happens in capacitors.

 

[gracy]

I know
$εo$=$\frac{σ}{E}$
Does the below formula exist
$ε$=$\frac{σ}{E}$

 

[gracy]

Susceptibility ,polarization when external field is applied ,permittivity ,relative permittivity are all these only for dielectrics not for conductors ?

 

[cnh1995]

I know
$εo$=$\frac{σ}{E}$
Does the below formula exist
$ε$=$\frac{σ}{E}$

Yes. ε=ε_oε_r. In case of vacuum, ε_r=1, so there's only ε_o in the formula.

 

[cnh1995]

Susceptibility ,polarization when external field is applied ,permittivity ,relative permittivity are all these only for dielectrics not for conductors ?

Conductors are extreme cases of permittivity and susceptibility.

 

[gracy]

Yes. ε=ε0εr. In case of vacuum, εr=1, so there's only εo in the formula.

$ε0$=$\frac{σ}{E}$
in case of vacuum
but in case of other materials we should apply the below formula
$ε$=$\frac{σ}{E}$
Right?

 

[cnh1995]

$ε0$=$\frac{σ}{E}$
in case of vacuum
but in case of other materials we should apply the below formula
$ε$=$\frac{σ}{E}$
Right?

Right.

 

[gracy]

Here (in the video)permittivity of vacuum is taken into consideration not of the slab ,why?

 

[cnh1995]

Here (in the video)permittivity of vacuum is taken into consideration not of the slab ,why?

Slab is metallic. E field inside it will be 0. The video is about how charge density on the metal plate will vary with the permittivity of the surroundings.

 

[gracy]

permittivity tells you how much charge is required(or permitted) to create the same electric field in different dielectrics.

How can I apply the above mentioned function of permittivity to the surrounding?