I How Do Dielectrics Influence Electric Flux Density in Electromagnetic Theory?

[bboo123]

In the 7th edition of the book "Elements of Electromagnetics by Matthew N. O. Sadiku"

On page 190 the author goes on to say:

"We now consider the case in which the dielectric region contains free charge.
If $\rho_v$ is the volume density of free charge, the total volume charge density $\rho_t$ is given by:
$$\rho_t = \rho_v + \rho_{pv} = \nabla.\epsilon_0E$$ (Where $\rho_{pv}$ is the volume charge density due to polarization of the dielectric.)

Hence,
$$\rho_v = \nabla.\epsilon_0E - \rho_{vp} = \nabla.(\epsilon_0E + P) = \nabla.D$$

We conclude that the net effect of the dielectric on the electric field $E$ is to increase $D$
inside it by the amount $P$. In other words, $\textbf{ the application of E to the dielectric material causes the flux density to be greater than it would be in free space. }$ "

Now my questions are:
1) I don't exactly get how did the author conclude the electric flux density increases by P from the last equation since E is definitely not the External electric field here, so it's wrong to compare it directly with the electric flux density in free space.
2) $\textbf{And this is my main question}$, if the dielectric did not have free charges, can we say that the electric flux density $D$ $\textbf{remains constant}$? i.e D is the same as it was in free space in the newly introduced dielectric.

 

[vanhees71]

A dielectric consists of positive and negative charges which are bound to each other, i.e., there are no charges which are quasi free to move within the material (as is the case for metals, which are conductors, because part of the electrons are not bound to specific atoms but delocalized over the entire material). If you now put the dielectric in an external electric field there's a force on the positive (negative) charges in (opposite to) the direction of the electric field. The external field is $\vec{E}_{\text{ext}}=\vec{D}/\epsilon_0$.

This induces dipoles pointing in the direction of $\vec{E}_{\text{ext}}$ and thus counteracting this field. At a point the induced dipole density (aka polarization) for not too large $\vec{E}_{\text{ext}}$ is proportional to the total electric field, i.e., the external field plus the field due to the dipoles. If the medium is isotropic the proportionality constant it's simply a scalar field, i.e.,
$$\vec{P}=\chi \epsilon_0 \vec{E},$$
where the factor $\epsilon_0$ is conventional.

Now the electrostatic potential due to the polarization is
$$\Phi_{\text{med}}(\vec{x})=-\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{P}(\vec{x}') \cdot \vec{\nabla} \frac{1}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}=+\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{P}(\vec{x}') \cdot \vec{\nabla}' \frac{1}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}=-\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \cdot \vec{P}(\vec{x}')}{4 \pi \epsilon|\vec{x}-\vec{x}'|}.$$
In the last step I've integrated by parts.

This means that the polarization is equivalent to a charge density
$$\rho_{\text{mat}}=-\vec{\nabla} \cdot \vec{P}.$$
So for the part of the induced electric field from the polarized medium you get
$$\vec{\nabla} \cdot \vec{E}_{\text{med}}=-\frac{1}{\epsilon_0} \vec{\nabla} \cdot \vec{P}.$$
But now
$$\vec{E}=\vec{E}_{\text{med}} + \vec{E}_{\text{ext}}$$
and thud
$$\vec{\nabla} \cdot \vec{E} = \frac{1}{\epsilon_0} (-\vec{\nabla} \cdot \vec{P}+\rho_{\text{ext}})$$
or
$$\vec{\nabla} (\epsilon_0 \vec{E}+\vec{P})=\rho_{\text{ext}}.$$
Now one introduces the auxilliary field $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$. Then
$$\vec{\nabla} \cdot \vec{D}=\rho_{\text{ext}}.$$
With the above linear ansatz for the polarization this means
$$\vec{D}=\epsilon_0 (1+\chi) \vec{E}=\epsilon_0 \epsilon_{\text{rel}} \vec{E},$$
where $\epsilon_{\text{rel}}$ is the relative permittivity of the medium. Since for usual materials $\vec{P}$ is in the direction of $\vec{E}$ (as made plausible by the above heuristic model of a dielectric) you have $\chi>0$ and thus $\epsilon_{\text{rel}}>1$.

Thus to get the same electric field inside the dielectric as in vacuum you need to make the external charges larger by a factor $\epsilon_{\text{rel}}$ to compensate for the counteracting electric field due to the polarization, $\vec{E}_{\text{mat}}$.

 

[bboo123]

A dielectric consists of positive and negative charges which are bound to each other, i.e., there are no charges which are quasi free to move within the material (as is the case for metals, which are conductors, because part of the electrons are not bound to specific atoms but delocalized over the entire material). If you now put the dielectric in an external electric field there's a force on the positive (negative) charges in (opposite to) the direction of the electric field. The external field is $\vec{E}_{\text{ext}}=\vec{D}/\epsilon_0$.

This induces dipoles pointing in the direction of $\vec{E}_{\text{ext}}$ and thus counteracting this field. At a point the induced dipole density (aka polarization) for not too large $\vec{E}_{\text{ext}}$ is proportional to the total electric field, i.e., the external field plus the field due to the dipoles. If the medium is isotropic the proportionality constant it's simply a scalar field, i.e.,
$$\vec{P}=\chi \epsilon_0 \vec{E},$$
where the factor $\epsilon_0$ is conventional.

Now the electrostatic potential due to the polarization is
$$\Phi_{\text{med}}(\vec{x})=-\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{P}(\vec{x}') \cdot \vec{\nabla} \frac{1}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}=+\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{P}(\vec{x}') \cdot \vec{\nabla}' \frac{1}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}=-\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \cdot \vec{P}(\vec{x}')}{4 \pi \epsilon|\vec{x}-\vec{x}'|}.$$
In the last step I've integrated by parts.

This means that the polarization is equivalent to a charge density
$$\rho_{\text{mat}}=-\vec{\nabla} \cdot \vec{P}.$$
So for the part of the induced electric field from the polarized medium you get
$$\vec{\nabla} \cdot \vec{E}_{\text{med}}=-\frac{1}{\epsilon_0} \vec{\nabla} \cdot \vec{P}.$$
But now
$$\vec{E}=\vec{E}_{\text{med}} + \vec{E}_{\text{ext}}$$
and thud
$$\vec{\nabla} \cdot \vec{E} = \frac{1}{\epsilon_0} (-\vec{\nabla} \cdot \vec{P}+\rho_{\text{ext}})$$
or
$$\vec{\nabla} (\epsilon_0 \vec{E}+\vec{P})=\rho_{\text{ext}}.$$
Now one introduces the auxilliary field $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$. Then
$$\vec{\nabla} \cdot \vec{D}=\rho_{\text{ext}}.$$
With the above linear ansatz for the polarization this means
$$\vec{D}=\epsilon_0 (1+\chi) \vec{E}=\epsilon_0 \epsilon_{\text{rel}} \vec{E},$$
where $\epsilon_{\text{rel}}$ is the relative permittivity of the medium. Since for usual materials $\vec{P}$ is in the direction of $\vec{E}$ (as made plausible by the above heuristic model of a dielectric) you have $\chi>0$ and thus $\epsilon_{\text{rel}}>1$.

Thus to get the same electric field inside the dielectric as in vacuum you need to make the external charges larger by a factor $\epsilon_{\text{rel}}$ to compensate for the counteracting electric field due to the polarization, $\vec{E}_{\text{mat}}$.

Thank you for the reply!

My main question was regarding the Electric Flux Density $D$.
So from what I have understood,

For an external electric field $E_0$, let us consider a point O,

D at O for the case of free space will be $\epsilon_0E_0$
Now, let us introduce an arbitrary dielectric medium at point O with a relative permittivity of $\epsilon_r$.
So now the D at point O in space will be:
$$D = \epsilon_r\epsilon_0 E_{dielectric}$$
$$\mbox{But we know, } E_{dielectric} = E_0/\epsilon_r$$
$$\mbox{Hence, } D = \epsilon_0E_0$$
Which is the same as free space. So doesn't this contradict the author who went on to say that the Electric flux density in a dielectric is $\vec{P}$ more than the Electric flux density in Free Space?

 

[vanhees71]

That would indeed be wrong, because, as you write
$$\vec{D}=\epsilon_0 \vec{E}+\vec{P}=\epsilon_0 \vec{E}_0.$$

 

[bboo123]

That would indeed be wrong, because, as you write
$$\vec{D}=\epsilon_0 \vec{E}+\vec{P}=\epsilon_0 \vec{E}_0.$$

I am not saying it is always true, but for the specific case when the dielectric is neutral and experiences linear polarization, it is true right?

 

[rude man]

I am not saying it is always true, but for the specific case when the dielectric is neutral and experiences linear polarization, it is true right?

Given an E field in vacuum, $\bf D = \epsilon_0 \bf E$.
Inserting a dielectric of permittivity $\epsilon$ increases $\bf D$ to $\bf D = \epsilon \bf E$. $\epsilon > \epsilon_0$ in general. I am assuming an isotropic dielectric.