Equation of dielectric polarization & net electric field

[OnAHyperbola]

In a linear isotropic homogeneous dielectric,

P= χε₀*E,

where E is the net electric field from the effects of both free and bound charges.

But E=(σ_free—P)/ε₀

So E itself depends on P.

My concern is, isn't this circular? How can the polarization density depend on a field that itself depends on it? Shouldn't it depend on the original field between the conducting plates, E₀, given by

E₀=σ_free/ε₀ ?

 

[Charles Link]

At a surface boundary, you get polarization surface polarization charge density ## \sigma_p= \vec{P} \cdot \hat{n} ## where ## \hat{n} ## is the outward pointing unit vector normal to the surface. This is a result of the equation ## -\nabla \cdot P=\rho_p ## for polarization charge density. The surface charge density that arises depends on P which depends on the internal E and the surface charges that arise can affect the internal E. Yes, it is a self-consistent mathematics that is needed to solve this. For some simple geometries, ( in cases where you get a uniform ## P ## from a uniform applied field ##E_o ##), ## E_i=E_o-(C)(P)/\epsilon_o ## where ## E_o ## is the applied field and ## E_i ## is the field in the material, with C being a factor that depends upon the geometry. Since ## P=\chi \epsilon_o E_i ##, (using mks units), ## E_i ## can be found in terms of ## E_o ## and ## \chi ##, (simple algebraic solution), if the geometric factor C is known. For a plane slab C=1, and for a sphere C=1/3. (Not sure what letter the textbooks use for this factor C in electrostatics. For the analogous problem in magnetostatics, they usually use the letter D.) And note to prove the 1/3 factor for a sphere requires some rather complex mathematics involving Legendre Polynomials. It's a very useful result for problems involving dielectric spheres. The plane slab factor (C=1) is much more readily computed and can be calculated by using Gauss's law. For the plane slab, ## E_i=E_o/(1+\chi) ## which is essentially a screening factor of the electric field inside the material given by the normalized dielectric constant ## \epsilon/\epsilon_o =1+\chi ##. ..editing... to understand these concepts in more detail, I would suggest working the problem of the case of fixed uniform polarization ## P ## of on a material that occurs spontaneously without any externally applied field ## E_o ## . You can do this for a plane slab or for a sphere. Compute the electric field that arises in the material as a result of the surface polarization charge from the polarization ## P ##. (The polarization ## P ## has no local effect of the electric field at that point. The only sources of electric field is the surface polarization charge density that arises at the boundaries. And assume in this case the the electric field that arises does not affect the polarization ## P ##. This is a simpler case than that of a material that has ## P=\epsilon_o \chi E ## , but a good one to illustrate the concept of surface polarization charge.)

 

[Delta2]

Well, one should be prepared for circular dependencies when studying electromagnetism (afterall its the circular dependency between electric and magnetic field that makes the magic of an EM wave happen) however here its not exactly circular

It is ##E=E_0-\frac{P}{\epsilon_0}## or ##E_0-E=\frac{P}{\epsilon_0}## which basically tell us that the Polarization depends on the initial and final value of E-field, it is the difference between them as we intuitively understand it pretty much...

 

[Charles Link]

Just an additional comment: The self-consistent calculation such as this one is a little tricky when you first see it, but the algebra is reasonably straightforward, at least for cases where the final result is a uniform field ## E_i ## and uniform polarization ## P ##. The mathematics for self-consistent calculations can get quite cumbersome in cases where the final result is a non-uniform field with a non-uniform polarization. Things would be very simple if the applied external field ## E_o ## were the field ## E_i ## inside the material as well, but oftentimes that is not the case.