There is exactly the same question on StackExchange but I don't find it rigorous or even right per se.
How flux density increases in the dielectric medium?
In the 7th edition of the book "Elements of Electromagnetics by Matthew N. O. Sadiku"
On page 190 the author goes on to say:
"We now consider the case in which the dielectric region contains free charge. If $\rho_v$ is the volume density of free charge, the total volume charge density $\rho_t$ is given by: $$\rho_t = \rho_v + \rho_{pv} = \nabla.\epsilon_0E$$ (Where $\rho_{pv}$ is the volume charge density due to polarization of the dielectric.)
Hence, $$\rho_v = \nabla.\epsilon_0E - \rho_{vp} = \nabla.(\epsilon_0E + P) = \nabla.D$$
We conclude that the net effect of the dielectric on the electric field $E$ is to increase $D$ inside it by the amount $P$. In other words, the application of $E$ to the dielectric material causes the flux density to be greater than it would be in free space. "
Now my questions are:
- I don't exactly get how did the author conclude the electric flux density increases by P from the last equation since E is definitely not the External electric field here, so it's wrong to compare it directly with the electric flux density in free space.
- $\textbf{And this is my main question}$, if the dielectric did not have free charges, can we say that the electric flux density $D$ $\textbf{remains constant}$?
