• pbsoftmml
In summary, the formula for the electric displacement field and capacitors can be derived by considering the charge density due to point charges and dipoles. This involves introducing the concept of polarization, which is the dipole-per-volume field, and defining a new field D that takes into account both polarization and free charge. It should be noted that D is a mathematical construction and is not a measurable physical quantity, while the real physical parameters are the electric field E and the polarization P. This formula can also be applied to the case of a parallel plate capacitor using Gauss' law.

#### 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!

#### Attachments

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} ##.

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• anorlunda and Cryo
pbsoftmml said:
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.

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?

Cryo said:
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?

pbsoftmml said:
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.

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See paragraph 4 of my post.
Oh sorry that I did not read through it.
Thank you!

pbsoftmml said:
Oh sorry that I did not read through it.
Thank you!
Be sure to see the Edit that I did to post 6.

pbsoftmml said:
How was the capacitance formula derived then?

## 1. How was the formula for electromagnetic force derived?

The formula for electromagnetic force, also known as Coulomb's law, was derived by French physicist Charles-Augustin de Coulomb in 1785. He conducted experiments on the force between charged particles and found that it was directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This relationship is expressed in the formula F = kq1q2/r2, where F is the force, q1 and q2 are the charges, r is the distance between them, and k is a constant known as the Coulomb constant.

## 2. What is the significance of the constant k in the electromagnetic force formula?

The constant k in the formula for electromagnetic force, also known as the Coulomb constant, represents the permittivity of free space. It is a fundamental constant that determines the strength of the force between two charged particles. Its value is approximately 8.99 x 109 N·m2/C2, and it is the same for all charged particles in the universe.

## 3. Can the formula for electromagnetic force be applied to all types of charged particles?

Yes, the formula for electromagnetic force can be applied to all types of charged particles, including protons, electrons, and ions. This is because the force between two charged particles is solely determined by their charges and the distance between them, as described by Coulomb's law.

## 4. How does the formula for electromagnetic force relate to the concept of electric fields?

The formula for electromagnetic force is closely related to the concept of electric fields. Electric fields are created by charged particles and can be thought of as a field of force that extends through space. The strength of the electric field at a certain point is determined by the force that would be exerted on a test charge placed at that point, as given by the formula E = F/q, where E is the electric field strength and q is the test charge. This relationship shows that the force between two charged particles is a result of the electric field created by those particles.

## 5. Are there any other formulas related to electromagnetic force?

Yes, there are other formulas related to electromagnetic force, such as Gauss's law, Ampere's law, and Faraday's law. These laws describe the behavior of electric and magnetic fields and how they are affected by charges and currents. By combining these laws with Coulomb's law, we can derive more complex formulas for electromagnetic interactions, such as the equations for capacitance and inductance.