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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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2 comment threads

This questions need more introduction (4 comments)
Where are you getting the $\sigma da = \rho d\tau$ equation? As you allude to, $da$ is referring to a... (4 comments)

Comments on How electric displacement is proportional to surface charge?

This questions need more introduction
Trilarion‭ wrote 9 months ago:

You throw the reader right into the action. That may or may not work well. It would be better to first state what you want to achieve, then how you think you can solve the problem and third your approach and what it yielded. As it is, one has to guess a lot why you did what you did here.

deleted user wrote 9 months ago:

I was just "playing" with equations randomly, what I can do with them. Latter, I found them. I wasn’t solving any problem. When I saw the first equation, I thought i can find another relation then I got it. I wasn’t expecting anything, I was just "playing".

Trilarion‭ wrote 9 months ago:

I really have problems understanding why you posted the question and the answer. I think it likely is not very useful to others because they as well will not understand why you are trying to achieve and where the problem is.

deleted user wrote 9 months ago:

It was mostly about "what displacement has to do with free surface charges?". That's what I think by proportionality.

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