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Comments on How electric displacement is proportional to surface charge?

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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)
Where are you getting the $\sigma da = \rho d\tau$ equation? As you allude to, $da$ is referring to a...
Derek Elkins‭ wrote over 2 years ago

Where are you getting the $\sigma da = \rho d\tau$ equation? As you allude to, $da$ is referring to an area element and $d\tau$ to a volume element, so it's not clear what it means to equate these expressions.

deleted user wrote over 2 years ago

That's written in Griffiths book, he used tilde to say that they are equal, and he used over and over again to find lots of equation.

And I have found that my derivation is correct cause he wrote the same equation 3-4 pages later. So I WILL edit my question latter

Derek Elkins‭ wrote over 2 years ago

Just because you get the correct result doesn't mean that your derivation is correct. At the time of writing the question you seemed uncertain as to how/why we could go from $\int_V \rho_f d\tau$ to $\oint_S \sigma_f da$. The way I read your concern(admittedly with a lot of guessing on my part to fill in blanks) was that you had $\int_V \rho_f d\tau$ and you had $\rho_f d\tau = \sigma_f da$ so naively one might think that you could replace $\rho_f d\tau$ with $\sigma_f da$ getting $\int_V \sigma_f da$, but that doesn't make any sense so you just changed the $\int_V$ to $\oint_S$ to get a meaningful expression. If you haven't already, it seems worthwhile to articulate to yourself clearly what is actually going on.

deleted user wrote about 2 years ago

$$dq=\rho d\tau$$ $$q=\int_V \rho d\tau$$ $$dq=\sigma da$$ $$q=\int_S \sigma da$$ I am not interested in further discussion with you who can't understand simple things so why should I discuss with them. (Just waste of my time :|). Unless prove me wrong.