# Post History

##
**#3: Post edited**

- Let's start with displacement field equation
- $$\vec D = \epsilon_0 \vec E + \vec P$$
- We know that $$-\vec \nabla \cdot \vec P = \rho_b$$
- Here $\rho_b$ is surface charge density.
- $$\implies -\int_V (\vec \nabla \cdot \vec P)\mathrm d\tau=\int_V \rho_b \mathrm d\tau$$
- $$\implies \vec P \cdot \hat n = \sigma_b$$
- The OP found $\vec D = \sigma_f \cdot \hat n$
- $\sigma_b$ is used for surface charge density while $\sigma_f$ is used for free surface charge density if you look at the displacement field equation then you will get
- $$\implies\sigma_f = \epsilon_0 \vec E+\sigma_b$$
- $$\implies\vec E = \frac{\sigma_f-\sigma_b}{\epsilon_0}$$
- $$=\frac{\sigma}{\epsilon_0}$$
- >The free charge density serves as a useful simplification in Gauss's law for electricity
- > Take a stationary atom and apply an electric field externally to the electron then the free particles align themselves with the field and create an electric dipole $\vec p = q\vec d$ where d is distance between those free particles.
- >If you take a look at electric field generally then you will get the same expression $\vec E = \frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r^2}da\\\\ \implies \vec E = \frac{\sigma}{\epsilon_0}$ so both equation agrees on each other.
- >So polarization is **surface charge density** and electric displacement is **free surface charge density** and direction of polarization and electric displacement depends on electric field.
- Simply, free particles being moved for electric and field and attraction and distraction of other particles
- References :
- 1 : https://en.wikipedia.org/wiki/Charge_density
~~2 : https://www.youtube.com/watch?v=-NjJiQN_cyA~~

- Let's start with displacement field equation
- $$\vec D = \epsilon_0 \vec E + \vec P$$
- We know that $$-\vec \nabla \cdot \vec P = \rho_b$$
- Here $\rho_b$ is surface charge density.
- $$\implies -\int_V (\vec \nabla \cdot \vec P)\mathrm d\tau=\int_V \rho_b \mathrm d\tau$$
- $$\implies \vec P \cdot \hat n = \sigma_b$$
- The OP found $\vec D = \sigma_f \cdot \hat n$
- $\sigma_b$ is used for surface charge density while $\sigma_f$ is used for free surface charge density if you look at the displacement field equation then you will get
- $$\implies\sigma_f = \epsilon_0 \vec E+\sigma_b$$
- $$\implies\vec E = \frac{\sigma_f-\sigma_b}{\epsilon_0}$$
- $$=\frac{\sigma}{\epsilon_0}$$
- >The free charge density serves as a useful simplification in Gauss's law for electricity
- > Take a stationary atom and apply an electric field externally to the electron then the free particles align themselves with the field and create an electric dipole $\vec p = q\vec d$ where d is distance between those free particles.
- >If you take a look at electric field generally then you will get the same expression $\vec E = \frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r^2}da\\\\ \implies \vec E = \frac{\sigma}{\epsilon_0}$ so both equation agrees on each other.
- >So polarization is **surface charge density** and electric displacement is **free surface charge density** and direction of polarization and electric displacement depends on electric field.
- Simply, free particles being moved for electric and field and attraction and distraction of other particles
- References :
- 1 : https://en.wikipedia.org/wiki/Charge_density
- 2 :
**[Electromagnetism: Displacement Field and Gauss's Law (YT video)](**https://www.youtube.com/watch?v=-NjJiQN_cyA**)** **Further reading :****[What is electric displacement](https://qr.ae/pG62QB)**

##
**#2: Post edited**

- Let's start with displacement field equation
- $$\vec D = \epsilon_0 \vec E + \vec P$$
- We know that $$-\vec \nabla \cdot \vec P = \rho_b$$
- Here $\rho_b$ is surface charge density.
- $$\implies -\int_V (\vec \nabla \cdot \vec P)\mathrm d\tau=\int_V \rho_b \mathrm d\tau$$
- $$\implies \vec P \cdot \hat n = \sigma_b$$
- The OP found $\vec D = \sigma_f \cdot \hat n$
- $\sigma_b$ is used for surface charge density while $\sigma_f$ is used for free surface charge density if you look at the displacement field equation then you will get
- $$\implies\sigma_f = \epsilon_0 \vec E+\sigma_b$$
- $$\implies\vec E = \frac{\sigma_f-\sigma_b}{\epsilon_0}$$
- $$=\frac{\sigma}{\epsilon_0}$$
- >The free charge density serves as a useful simplification in Gauss's law for electricity
- > Take a stationary atom and apply an electric field externally to the electron then the free particles align themselves with the field and create an electric dipole $\vec p = q\vec d$ where d is distance between those free particles.
- >If you take a look at electric field generally then you will get the same expression $\vec E = \frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r^2}da\\\\ \implies \vec E = \frac{\sigma}{\epsilon_0}$ so both equation agrees on each other.
- >So polarization is **surface charge density** and electric displacement is **free surface charge density** and direction of polarization and electric displacement depends on electric field.
~~Simply, free particles being moved for electric and field and attraction and distraction of other particles~~

- Let's start with displacement field equation
- $$\vec D = \epsilon_0 \vec E + \vec P$$
- We know that $$-\vec \nabla \cdot \vec P = \rho_b$$
- Here $\rho_b$ is surface charge density.
- $$\implies -\int_V (\vec \nabla \cdot \vec P)\mathrm d\tau=\int_V \rho_b \mathrm d\tau$$
- $$\implies \vec P \cdot \hat n = \sigma_b$$
- The OP found $\vec D = \sigma_f \cdot \hat n$
- $$\implies\sigma_f = \epsilon_0 \vec E+\sigma_b$$
- $$\implies\vec E = \frac{\sigma_f-\sigma_b}{\epsilon_0}$$
- $$=\frac{\sigma}{\epsilon_0}$$
- >The free charge density serves as a useful simplification in Gauss's law for electricity
**References :****1 : https://en.wikipedia.org/wiki/Charge_density****2 : https://www.youtube.com/watch?v=-NjJiQN_cyA**