Understanding Dielectric Polarisation: Transformation of Dipoles and Charges

[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?

 

[cnh1995]

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

There will be field lines from +ve charge density towards -ve charge density "through" the surrounding. It is similar to what happens inside a capacitor.

 

[gracy]

You mean some charges must be there to produce that "applied electric field".And then permittivity of surrounding will come into picture.The fields shown in the video have gone through all of these.

 

[cnh1995]

You mean some charges must be there to produce that "applied electric field".And then permittivity of surrounding will come into picture.The fields shown in the video have gone through all of these.

Yes. The charge densities act as a dipole. So, there will be field lines in the surroundings, joining the two charge densities.

 

[gracy]

permittivity tells you how "strong"(may not be the correct word) a dielectric is w.r.t vacuum( free space)

But it is permittivity of free space that tells us how "strong"(may not be the correct word) a dielectric is w.r.t vacuum( free space) if we look at permittivity of other materials how it differs from their susceptibility?

 

[nasu]

Both of them tell something about how "strong" is the dielectric but this is just a "metaphoric" way to describe them. You spend way too much time with matter of little relevance and questions expressed in somehow confusing terms.
The equations tells it all without any room for misinterpretation: susceptibility is used to write the relationship between external field and polarization whereas relative permitivity (or dielectric strength or dielectric constant) for the relationship between the external field and the net field in the dielectric. They both depend on the same property of the material: the ability to "produce" electric dipoles in the presence of a field. None of them has more (or less) meaning than the other. If you know one you can get the other.

The vacuum permitivity you can leave out of this. It is more like a matter of units and not really o property of a "material" in the same sense as the dielectric constant.

 

[cnh1995]

But it is permittivity of free space that tells us how

No. I said it is the permittivity of the material, not free space.

 

[gracy]

permittivity gives the relation between net field and the original field

You meant relative permittivity?

 

[cnh1995]

You meant relative permittivity?

I meant actual permittivity(ε) of the material(ε_oε_r).

 

[gracy]

But it's relative permittivity which is also called dsielectric constant that's gives the relation between net field and the original field
$K$=$\frac{E_O}{E}$

 

[cnh1995]

But it's relative permittivity which is also called dsielectric constant that's gives the relation between net field and the original field
$K$=$\frac{E_O}{E}$

Right.

 

[gracy]

That's what I asked in post #42

 

[cnh1995]

That's what I asked in post #42

Yeah right. I meant the concept of permittivity (relative permittivity is a part of that) compared to the concept of susceptibility.