How was this formula derived? (Electromagnetic)

[pbsoftmml]

I recently learned about Electric displacement field and capacitors, and I have a question that how was the formula derived shown below (blue circle part)?

Thanks!

 

[Charles Link]

These equations come from $\nabla \cdot E=\frac{\rho_{total}}{\epsilon_o}$ along with $-\nabla \cdot P=\rho_p$. Defining $D=\epsilon_o E+P$, we can take the divergence of both sides: $\nabla \cdot D=\epsilon_o \nabla \cdot E+\nabla \cdot P=\rho_{total}-\rho_p=\rho_{free}$. $\\$ Meanwhile, polarization $P=\epsilon_o \chi E$, where $\chi$ is called the dielectric susceptibility, so that $D=\epsilon_o (1+\chi ) E=\epsilon E=\epsilon_o \epsilon_r E$. The constant $K=\epsilon_r =1+\chi$. We can also write $P=(K-1)\epsilon_o E$. $\\$ It should be noted that $D$ is a mathematical construction, and not any real measurable physical quantity. There is no instrument that is able to distinguish polarization charge from free charge. The electric field $E$ is the real physical parameter here, along with $P$ which is the polarization (number of dipole moments per unit volume) in the material. $\\$ For the case of the parallel plate capacitor, you can use Gauss' law in the form $\int D \cdot dA=Q_{free}=\sigma_{free}A$ where the Gaussian pillbox has one face in the dielectric material between the plates, and the other face is outside of the capacitor. (Putting it in the conductive plate is also possible, but not good practice, because there is no guarantee that $D=0$ in the conductor, just because $E=0$ in the conductor). This gives $K \epsilon_o E A=\sigma_{free}A$, and $V=Ed$, and also $C=\frac{Q}{V}$. $\\$ Introducing $D$ can sometimes make the mathematics a little quicker, but the same calculations can always be done working with Gauss' law $\int E \cdot dA=\frac{Q_{total}}{\epsilon_o}$, and $-\nabla \cdot P=\rho_p$, which gives a Gauss' law for $P$ in the form $\int P \cdot dA=-Q_p$. One important relation that follows from this last result is that the polarization surface charge density $\sigma_p=P \cdot \hat{n}$.

 

[Cryo]

I recently learned about Electric displacement field and capacitors, and I have a question that how was the formula derived shown below (blue circle part)?
View attachment 230188
Thanks!

There is a way to derive this relationship starting from dipoles. First you start with stating the charge density due to point-charge:

$\rho\left(\vec{r}\right)=q\delta^{(3)}\left(\vec{r}-\vec{r_0}\right)$, where $\delta^{(3)}$ is the 3D delta function, $q$ is the charge and $\vec{r}_0$

is the position of the charge. This formula is self-evident, it is indeed non-zero only at one point, and if you integrate over it you get the actual charge.

Based on this, by considering two closely spaced opposite charges (of same magnitude), you can derive the charge density due to point-electric dipole:

$\rho_p=-\vec{\nabla}.\vec{p}\delta^{(3)}\left(\vec{r}-\vec{r}_p\right)$ where $\vec{p}$ is the dipole moment and $\vec{r}_p$

is the position of the dipole. Next we can introduce the density of the dipoles, i.e. the field that gives the dipole-per-volume:

$\vec{P}=\frac{1}{Vol}\int_{Vol}d^3 r_p\: \vec{p}(\vec{r}_p)\delta^{(3)}\left(\vec{r}-\vec{r}_p\right)$

where $Vol$ is the volume we are considering. Finally we can choose the separate the charge density into 'free', and charge density bound in dipoles:

$\vec{\nabla}.\varepsilon_0\vec{E}=\rho=\rho_P+\rho_{free}=-\vec{\nabla}.\vec{P}+\rho_{free}$
$\vec{\nabla}.\left(\varepsilon_0\vec{E}+\vec{P}\right)=\rho_{free}$

Where $\vec{P}$ is still the dipole density, which is commonly called polarization. Finally we define field $\vec{D}=\varepsilon_0\vec{E}+\vec{P}$ and we are done. By considering the current due to oscillating point-dipole, you can also introduce $\vec{P}$ into curl-H equation.

 

[pbsoftmml]

These equations come from $\nabla \cdot E=\frac{\rho_{total}}{\epsilon_o}$ along with $-\nabla \cdot P=\rho_p$. Defining $D=\epsilon_o E+P$, we can take the divergence of both sides: $\nabla \cdot D=\epsilon_o \nabla \cdot E+\nabla \cdot P=\rho_{total}-\rho_p=\rho_{free}$. $\\$ Meanwhile, polarization $P=\epsilon_o \chi E$, where $\chi$ is called the dielectric susceptibility, so that $D=\epsilon_o (1+\chi ) E=\epsilon E=\epsilon_o \epsilon_r E$. The constant $K=\epsilon_r =1+\chi$. We can also write $P=(K-1)\epsilon_o E$. $\\$ It should be noted that $D$ is a mathematical construction, and not any real measurable physical quantity. There is no instrument that is able to distinguish polarization charge from free charge. The electric field $E$ is the real physical parameter here, along with $P$ which is the polarization (number of dipole moments per unit volume) in the material. $\\$ For the case of the parallel plate capacitor, you can use Gauss' law in the form $\int D \cdot dA=Q_{free}=\sigma_{free}A$ where the Gaussian pillbox has one face in the dielectric material between the plates, and the other face is outside of the capacitor. (Putting it in the conductive plate is also possible, but not good practice, because there is no guarantee that $D=0$ in the conductor, just because $E=0$ in the conductor). This gives $K \epsilon_o E A=\sigma_{free}A$, and $V=Ed$, and also $C=\frac{Q}{V}$. $\\$ Introducing $D$ can sometimes make the mathematics a little quicker, but the same calculations can always be done working with Gauss' law $\int E \cdot dA=\frac{Q_{total}}{\epsilon_o}$, and $-\nabla \cdot P=\rho_p$, which gives a Gauss' law for $P$ in the form $\int P \cdot dA=-Q_p$. One important relation that follows from this last result is that the polarization surface charge density $\sigma_p=P \cdot \hat{n}$.

How was the capacitance formula derived then?

 

[pbsoftmml]

There is a way to derive this relationship starting from dipoles. First you start with stating the charge density due to point-charge:

$\rho\left(\vec{r}\right)=q\delta^{(3)}\left(\vec{r}-\vec{r_0}\right)$, where $\delta^{(3)}$ is the 3D delta function, $q$ is the charge and $\vec{r}_0$

is the position of the charge. This formula is self-evident, it is indeed non-zero only at one point, and if you integrate over it you get the actual charge.

Based on this, by considering two closely spaced opposite charges (of same magnitude), you can derive the charge density due to point-electric dipole:

$\rho_p=-\vec{\nabla}.\vec{p}\delta^{(3)}\left(\vec{r}-\vec{r}_p\right)$ where $\vec{p}$ is the dipole moment and $\vec{r}_p$

is the position of the dipole. Next we can introduce the density of the dipoles, i.e. the field that gives the dipole-per-volume:

$\vec{P}=\frac{1}{Vol}\int_{Vol}d^3 r_p\: \vec{p}(\vec{r}_p)\delta^{(3)}\left(\vec{r}-\vec{r}_p\right)$

where $Vol$ is the volume we are considering. Finally we can choose the separate the charge density into 'free', and charge density bound in dipoles:

$\vec{\nabla}.\varepsilon_0\vec{E}=\rho=\rho_P+\rho_{free}=-\vec{\nabla}.\vec{P}+\rho_{free}$
$\vec{\nabla}.\left(\varepsilon_0\vec{E}+\vec{P}\right)=\rho_{free}$

Where $\vec{P}$ is still the dipole density, which is commonly called polarization. Finally we define field $\vec{D}=\varepsilon_0\vec{E}+\vec{P}$ and we are done. By considering the current due to oscillating point-dipole, you can also introduce $\vec{P}$ into curl-H equation.

How was the capacitance formula derived then?

 

[Charles Link]

How was the capacitance formula derived then?

See paragraph 4 of my post. $\\$ Edit: They made a mistake in their last equation of the capacitor=it should be an $E$ rather than a $D$ in the numerator, or else a $D$ in the numerator but no $K \epsilon_o$ in front.

 

[pbsoftmml]

See paragraph 4 of my post.

Oh sorry that I did not read through it.
Thank you!

 

[Charles Link]

Oh sorry that I did not read through it.
Thank you!

Be sure to see the Edit that I did to post 6.

 

[Cryo]

How was the capacitance formula derived then?

:-) Sorry, I misread your question. I thought you were asking about the relationship between $\vec{P}$ and $\vec{D}$. Charles Link already did what you were actually asking for.