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So, we started from $$\oint \vec{E}d\vec{A}=\frac{Q}{\epsilon_0}$$ And used an electric dipole setup with $\vec{p}=q\vec{l}$ and $\vec{P}=\frac{\sum\vec{p}}{V}$, and reached the desired result: $$\oint \vec{D}d\vec{A}=Q_{real}$$ But I was thinking about an easier way. Formally: $$Q_{polarization}=\frac{\vec{p}}{\vec{l}}=\frac{\vec{P}V}{\vec{l}}=\vec{P}\vec{A}$$ But it's not correct mathematically. I think I should do some infinitesimal calculations to get $Q_{p}=\int \vec{P} d \vec{A}$, but how should I do it?

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  • $\begingroup$ Well one problem is that you cannot divide by a vector... $\endgroup$ – Aaron Stevens Jul 2 '18 at 13:17
  • $\begingroup$ @AaronStevens As I wrote, it's a formal, mathematically not correct derivation. $\endgroup$ – 545941st user Jul 2 '18 at 13:20
  • $\begingroup$ Maybe you could try rewriting as $q l\hat l$? $\endgroup$ – Aaron Stevens Jul 2 '18 at 13:29
  • $\begingroup$ Suggestion to the post (v2): Replace the word Maxwell's equations with the word Gauss' law in various places. $\endgroup$ – Qmechanic Jul 2 '18 at 17:28

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