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I was in doubt about electric displacement, after some time I tried to find the unit of $D$ which is $Cm^{-2}$. Why?

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    $\begingroup$ Because of how $\mathbf{D}$ is defined in SI units. Don't think of it as a surface charge density. Think $q/r^2$. $\endgroup$ – G. Smith Mar 20 at 4:39
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    $\begingroup$ But what is the physical significance of electric displacement? $\endgroup$ – Nikhil Pathak Mar 20 at 5:03
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    $\begingroup$ An $\mathbf{E}$ field applied to matter (as opposed to vacuum) polarizes the atoms in the matter. This creates an additional electric field. The $\mathbf{D}$ field takes both into account. $\endgroup$ – G. Smith Mar 20 at 5:12
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    $\begingroup$ Is this $E$ external or internal electric field produced due to Polarization?And how $D$ field takes both into account? $\endgroup$ – Nikhil Pathak Mar 20 at 5:27
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    $\begingroup$ Actually, I think it is $\mathbf{E}$ that includes both. $\mathbf{D}$ is the part that doesn’t include the field from the polarized atoms. Sorry for the confusion. I am better at thinking about fields in vacuum than in matter. $\endgroup$ – G. Smith Mar 20 at 5:31

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