I was in doubt about electric displacement, after some time I tried to find the unit of $D$ which is $Cm^{-2}$. Why?
Asked
Active
Viewed 217 times
0
Qmechanic
- 201,751
-
1Because 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 '20 at 04:39
-
1But what is the physical significance of electric displacement? – Nikhil Pathak Mar 20 '20 at 05:03
-
1An $\mathbf{E}$ 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 '20 at 05:12
-
1Is this $E$ external or internal electric field produced due to Polarization?And how $D$ field takes both into account? – Nikhil Pathak Mar 20 '20 at 05:27
-
1Actually, 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 '20 at 05:31
-
I am very confused.May you give me some reference from where I can learn about it? – Nikhil Pathak Mar 20 '20 at 05:34
-
Well, if you are unhappy with your textbook, I’m not sure that Wikipedia will seem more helpful. – G. Smith Mar 20 '20 at 05:36
-
The various answers to this PSE question may be useful. – G. Smith Mar 20 '20 at 05:39
-
$\nabla \cdot \vec D = \rho$ where $\rho$ is the volume charge density. – Farcher Mar 20 '20 at 06:30