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# How electric displacement is proportional to surface charge?

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\begin{alignat}{2} && \vec \nabla \cdot \vec D & = \rho_f \\ & \implies &\int_V \vec{\nabla} \cdot \vec D \mathrm d\tau & = \int_V \rho_f\ \mathrm d \tau \\ & \implies & \oint_S \vec D \cdot \mathrm d \vec a & = \oint_S \sigma_f \ da \end{alignat}

I used divergence theorem for the first term and $dq=\sigma da=\rho d\tau$ (integral changes for different values.(?)) for the second term. I was thinking I should use closed integral for the second term or not. Then I realized I was thinking of charge density in a specific field (we are ignoring outside charges) that's why I took a closed field.

Since both integrand are same we can take out both integral.

$$\vec D\cdot \hat n = \sigma_f$$

Here $\vec D \cdot d\vec a = \vec D \cdot \hat n \ da$ and $\sigma_f$ is used for free charges in an area. I know that $\vec D$ is electric displacement. But what wondered me here that is how electric displacement is proportional (equal?) to surface charge area. After deriving the equation what came to my mind that is perhaps my understanding of electric displacement is wrong.

when we put a charge conductor (which is uniformly charged) in a charge-less conductor the chargeless conductor feels some attraction toward electric field, that's what called electric displacement.

I think my understanding is completely wrong. So what is electric displacement? And is my derivation really correct? I had found a Wikipedia page, they didn't explain but wrote "further explanation needed" (they want someone else to re-edit that page).

While solving problem 4.18 (a) of Griffiths EM I found the similar equation where they found $$\int \vec D \cdot da=Q_{f_{enc}} \\ \implies \vec D A=\sigma A \\ \implies \vec D=\sigma$$ my approach was related also.

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This questions need more introduction (4 comments)

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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 :

What is electric displacement

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