Why don't we care about bound charges inside dielectric?

Question

I'm quoting Marek's answer from the question : Difference between electric field E and electric displacement field D

..materials have lots of internal charges you usually don't care about. You can get rid of them by introducing polarization P (which is the material's response to the applied E field). Then you can subtract the effect of internal charges and you'll obtain equations just for free charges

I couldn't figure out what are the exact effects he is talking about at first. After that I saw that maybe it is due to the fact that bound charges do not contribute to electric flux desnity field D outside the material (because flux line originates from bounded positive charge and terminates on bounded negative charge inside the dielectric, is it correct?). Is it one of the reasons that we are at large not interested in bound charges and have developed suitable mathematical ways to eliminate troubles caused by them? Are there any other reasons?

Thanks in advance.

Answer 1

We cannot control the bound charges, but we can control the free charges. Therefore, it makes sense that we would want to refer to something that only depends on that which we can control. This is why we introduce the electric displacement $$\mathbf D=\epsilon_0\mathbf E+\mathbf P$$ where the bound charge volume density $\rho_b$ relates to the polarization by $$\rho_b=-\nabla\cdot\mathbf P$$

On the other hand, by construction, the electric displacement only depends on the free charge density $\rho_f$$^*$ $$\nabla\cdot\mathbf D=\rho_f$$

So you can see why this is useful. We can't control and might not even know what the bound charge is doing in our system. But we control the free charge, so it makes much more sense to use $\mathbf D$ when thinking about electric fields in matter. This is especially true if we assume that the polarization is proportional to the field for linear dielectrics.

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$^*$Although if the curl of $\mathbf P$ is non-zero then we have $$\nabla\times\mathbf D=\nabla\times\mathbf P$$ and so $\mathbf D$ is not solely determined by the free charge.