Polarization of a di-electric due to an electric field

In summary, the bound charges inside a dielectric must sum up to zero due to the flux of charge entering and leaving the surface.
  • #1
if a di-electric was introduced to an electric field from the outside it would be polarized and the charge denisties would be as follows :
**volume bounded charge density ( ρv ) = - div ( P )
**surface bounded charge density ( ρs ) = P . n
where P is the polarization vector and n is the unit vector.

now if the di-electric was originally neutral , should all these bounded charges sum up to zero ?
( volume bounded charge density ρv + surface bounded charge density ρs = 0 ) ?


and what if the source of the electric field was coming from inside the di-electric ( as for example a free point charge ) would it still be the case ?
 
Engineering news on Phys.org
  • #2
now if the di-electric was originally neutral , should all these bounded charges sum up to zero ?

We can be certain of it because charge is conserved. We can assume that if polarization causes some distribution of charge then each negative region is balanced by another positive region.

Is that good enough? It wouldn't make any physical sense to create charges by some special configuration of electric fields.

If you want a more mathy answer then take a look at Gauss's law in integral form:

[tex]\oint\oint E \cdot dA = \frac {Q}{\epsilon_0}[/tex]

E is the electric field and Q is the total charge inside some closed surface over which the integration takes place. The polarization field is a direct function of E so there must be a proof of charge conservation somewhere in Gauss's law. I'll give it some more thought later.
 
  • #3
I think I have mathematical answer for you. Let's define flux:

[tex]\Phi = \oint\oint{\vec{V} \cdot \hat{n}}[/tex]

What this tells us is that some scalar(phi) is equal to the sum of the vectors pointing into and out of a closed surface. For an analogy, V could be the flow of water and the closed surface could be the outline of a lake. If more rivers are flowing into the lake than out, there will be a build up of water in the lake.

The electrical version is this (Gauss's law):

[tex]Q = \oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}}[/tex]

This tells us that the sum of electrical vectors pointing into and out of a surface is related to the charge inside the surface. If that charge starts as 0 then we have:

[tex]\oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}} = 0[/tex]

For linear dielectrics, we know what P is:

[tex]P = \epsilon_0\chi \vec{E} ; \vec{E} = P / (\epsilon_0\chi)[/tex]

So, we know right away that the bound surface charges must sum up to zero.

[tex]\sigma_s = P \cdot \hat{dA}[/tex]

[tex]\oint\oint{ \epsilon_0 \vec{E} \cdot \hat{dA}} = 0[/tex]

[tex]\oint\oint{ \epsilon_0 \left( P / (\epsilon_0\chi) \right) \cdot \hat{dA}} = 0[/tex]

[tex]\oint\oint{ \epsilon_0 \left( P \cdot \hat{dA} \right) \frac{1}{\epsilon_0\chi}} = 0[/tex]

[tex]\oint\oint{ \epsilon_0 \left( \sigma_s} \right) \frac{1}{\epsilon_0\chi}} = 0[/tex]

Now, the bound charges inside the dielectric are this:

[tex] \rho_v = -\nabla \vec{P}[/tex]

It is a property of vector fields that the following is true:

[tex]\oint\oint\oint \left( \nabla \vec{V} \right) = \Phi = \oint\oint \vec{V} \cdot \vec{dA} [/tex]

Therefore, the bound charges must sum up to zero because the flux going into the surface must be zero.

We could do all that again with a charge inside the surface and find the relationship between bound charges and the free internal charge. Also, I did this analysis for a linear dielectric. I'm not even sure what how to go about describing what would happen in a non-linear material.

Remember the simple answer? It was the one about how charge must be conserved so charged regions must balance out somehow. That answer is looking better and better don't you think?
 
  • #4
That was very helpful really.
yes now i understand , thanks very much Okefenokee .
 

What is polarization of a di-electric?

Polarization of a di-electric is a phenomenon in which the molecules of a non-conducting material, also known as a di-electric, become aligned in a specific direction when exposed to an electric field.

How does an electric field cause polarization?

When an electric field is applied to a di-electric material, the positive and negative charges within the molecules are separated, causing the molecules to become aligned in the direction of the field. This results in an overall dipole moment for the material.

What is the difference between polar and non-polar di-electric materials?

Polar di-electric materials have permanent dipole moments, meaning the alignment of their molecules remains even when no electric field is present. Non-polar di-electric materials do not have permanent dipole moments and their molecules only become aligned when exposed to an electric field.

What is the relationship between polarization and dielectric constant?

The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electric charge. Polarization of a di-electric material increases its dielectric constant, as the aligned molecules create a stronger electric field within the material.

How does polarization affect the behavior of a di-electric material?

Polarization changes the electrical properties of a di-electric material, including its capacitance, insulation properties, and ability to store charge. It also plays a crucial role in the functioning of electronic devices such as capacitors and transistors.

Suggested for: Polarization of a di-electric due to an electric field

Replies
20
Views
365
Replies
5
Views
1K
Replies
7
Views
536
Replies
19
Views
999
Replies
20
Views
2K
Replies
2
Views
703
Replies
6
Views
712
Replies
6
Views
800
Replies
18
Views
798
Replies
5
Views
1K
Back
Top