I want to know how this Result Is Derived.
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When electrifying the dielectric, the charges that comprise it begin to separate. So consider the work needed to shift the charge dq=d\sigma A (for example, such charge appears as bounded charges in the dielectric) by l. It equals to
\delta W = dq\cdot E\cdot l = d\sigma E\cdot Al = d\sigma EV.
Now, according to the Gauss flux theorem, the difference in the charge density by d\sigma between two small layers of the area A with the distance l between each other corresponds to the small increment of the \vec D through
dD = d\sigma V,
and thus, we get \delta W = E dD. Generally, only the projection of \vec D along the vector \vec E contributes to the work, so the general form would be the dot product \delta W = \vec Ed\vec D. Now note that the work done by external forces to electrify the dielectric entirely increases the electric field. So, having \vec D = \varepsilon\varepsilon_0\vec E and integrating, we get
u = W/V = \frac{\varepsilon\varepsilon_0 E^2}{2}.
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The electric displacement vector.
@eudaimon Please explain to StudyHere what is \vec D and what’s the difference between the electric displacement field and the electric field \vec E itself.
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@StudyHere, the vector \vec{D}, known as the electric displacement field, is a vector field associated with the presence of electric charges in dielectric materials, where polarization effects are significant. It represents the flux density of the electric field lines that are generated by free charges and induced charges (polarization charges) in a medium.
The electric displacement field \vec{D} is related to the electric field \vec{E} and the polarization \vec{P} of the material through the relationship:
\vec{D} = \varepsilon_0 \vec{E} + \vec{P},
where \varepsilon_0 is the permittivity of free space. In the presence of a dielectric material, this relationship becomes:
\vec{D} = \varepsilon \varepsilon_0 \vec{E},
where \varepsilon is the relative permittivity (also known as the dielectric constant) of the material, indicating how much the material can reduce the electric field within itself compared to the electric field in a vacuum.
The electric field \vec{E} represents the force per unit charge experienced by a small positive test charge placed in the field. It is not directly affected by the medium in which it propagates, and it is the field that directly influences the dynamics of charged particles.
In contrast, \vec{D} accounts for the material’s response to the electric field, including the alignment of dipoles within the material and the creation of bound charges at the boundaries. It is especially useful when dealing with Gauss’s law in media where polarization must be considered, as it simplifies the law to a form similar to that in vacuum:
\oint \vec{D} \cdot d\vec{A} = Q_{\text{free}},
where the integral is taken over a closed surface, and Q_{\text{free}} is the free charge enclosed by that surface.
In summary, while \vec{E} describes the force on charges, \vec{D} describes how that field is affected by the presence of a material, and it is particularly important when considering the storage and movement of energy within dielectric media.
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@eudaimon Explain StudyHere what are vectors \vec D and \vec P and how they are derived.
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Ah, @StudyHere, you seek to grasp the profound concepts of the electric field. Let us embark on this intellectual odyssey together.
The electric displacement field, denoted as \vec{D}, is a vector field that appears in Maxwell’s equations. It represents how an electric field \vec{E} influences the organization of electrical charges in a material medium, resulting in an electric polarization \vec{P}.
The polarization vector \vec{P}, on the other hand, is a measure of the separation of positive and negative charges within a material. It is defined as the electric dipole moment per unit volume of the material.
The relationship between \vec{D}, \vec{E}, and \vec{P} in a linear, isotropic, and non-conductive medium is given by:
\vec{D} = \varepsilon_0 \vec{E} + \vec{P}
Here, \varepsilon_0 is the permittivity of free space, a fundamental constant that characterizes the strength of the electric field in a vacuum.
To understand how these vectors are related, consider this: In the absence of any material medium, the electric displacement field \vec{D} is simply the product of the permittivity of free space \varepsilon_0 and the electric field \vec{E}. Introduce a material into the field, and the charges within this material will respond to \vec{E}, becoming polarized. This polarization, \vec{P}, modifies the original electric field, resulting in an adjusted electric displacement field \vec{D}.
Reflect upon this, @StudyHere: How might the presence of a medium change the way an electric field behaves? What does this imply about the relationship between \vec{E} and \vec{P}?
I’d add to Eudaimon’s response that the relation between \vec D, \vec P, and \vec E is derived through the Gauss’s law. In the well-known formula
\oint\vec E\cdot d\vec A = \frac{Q}{\varepsilon_0}.
The charge Q is understood as a sum of free charges Q_\text{free} and bound charges Q_\text{bound}.
The latter quantity is the sum of all polarized atoms/molecules that comprise the dielectric which acquire their own electric dipole moment \vec p as the external vector field \vec E increases. In other words, this linear dependency is expressed as \vec p =\beta\varepsilon_0\vec E, where \beta is called [molecular] polarizability.
Now, if we express Gauss’s law for the vector \vec P, which, as Eudaimon said, is the same as the dipole moment (\vec p\times\text{number of dipoles}) of a unit volume, we get
-\oint\vec P\cdot d\vec A = Q_\text{bound}
The minus sign appears because \vec p is accepted to be pointed from a negative charge to a positive charge, so for a positive vector \vec P flux (the vector is pointed from inside the closed surface) we should have negative charges inside the closed surface. So, plugging it to the initial Gauss’s law, we have
\oint (\varepsilon_0\vec E +\vec P)\cdot d\vec A = Q_\text{free}.
This is where we declare \vec D\equiv \varepsilon_0\vec E +\vec P as a new vector which depends only on free charges.
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