I found a lot of definitions of the electric displacement and none of them made sense to me. Some say it's the electric field in the dielectric, some say it's the density of free charges, and some say it's the flux of electric field due to free charges per unit area. What is the right definition? And are free charges the charges on the conductor that don't cancel the bound charges?
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2$\begingroup$ Answers to this may help: physics.stackexchange.com/questions/300741/… $\endgroup$– Andrew SteaneCommented Sep 13, 2022 at 15:36
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Most often, the expression $\epsilon_0 \mathbf E + \mathbf P$ (where $\mathbf P$ is polarisation) is taken as a definition. But since $\operatorname{div} \mathbf D = \rho$ (where $\rho$ is free charges density), this law also can be a base for the definition. (But this law does not fix $\operatorname{rot} \mathbf D$, so it should be defined additionally.) There is no difference here, and any of the two equations can be taken as a definition.
But there is also a more tricky level here. The point is in vacuum the electric field can be completely described by one vector: the electric field $\mathbf E$. The same picture takes place in a dielectric at the microscopic level. But we usually study a dielectric at the macroscopic level. And to do this, we need to average the microscopic electric field. It turns out that this can be done in two different ways, which give $\mathbf E$ and $\mathbf D$, respectively. And if we cut in a dielectric a cylindrical cavity with large radius and small height and measure the electic field in it (using a probe charge), the value will be equal to $\mathbf D$, and if we do the same in a cavity with small radius and large height, then $\mathbf E$ will be measured. These facts can also be taken as the definition of the corresponding values, and in this case their fundamental status equality is especially emphasized.
