# Why does the electric displacement vector $D$ have the same unit of charge density?

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

• Because of how $\mathbf{D}$ is defined in SI units. Don't think of it as a surface charge density. Think $q/r^2$. – G. Smith Mar 20 at 4:39
• But what is the physical significance of electric displacement? – Nikhil Pathak Mar 20 at 5:03
• 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. – G. Smith Mar 20 at 5:12
• Is this $E$ external or internal electric field produced due to Polarization?And how $D$ field takes both into account? – Nikhil Pathak Mar 20 at 5:27
• 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. – G. Smith Mar 20 at 5:31