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It is intuitively obvious that free charge is conserved in vaccum. But when is bound charge conserved? I tried to find this by formula:

The charge conservation law in microscopic view (Assuming there is zero displacement current):
$$ \nabla \cdot \vec{J} = - \frac{\partial \rho}{\partial t} $$ Assuming free charges remain free and bound charges remain bound:
$$ \nabla \cdot \vec{J_b} = - \frac{\partial \rho_b}{\partial t} $$ Decomposing the bound current yields:
$$ \nabla \cdot (\nabla \times \vec{M} + \frac{\partial \vec{P}}{\partial t}) = - \frac{\partial \rho_b}{\partial t} $$ Since divergence through curl is zero:
$$ \nabla \cdot \frac{\partial \vec{P}}{\partial t} = - \frac{\partial \rho_b}{\partial t} $$ Assuming polarization is continuous in space and time:
$$ \frac{\partial}{\partial t}(\nabla \cdot \vec{P}) = - \frac{\partial \rho_b}{\partial t} $$ Optionally, assuming polarization and bound charge are initially zero: $$ \nabla \cdot \vec{P} = -\rho_b $$

I can't think of any actual situation that demonstrates this. For example, in a capacitor, free charges turn bound, or vice versa, so this isn't a case.

Is there any actual situation that demonstrates this?

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closed as unclear what you're asking by Aaron Stevens, Jon Custer, user191954, Norbert Schuch, hyportnex Nov 19 '18 at 0:14

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You are asking for an example of when $\nabla\cdot\vec P=-\rho_b$? $\endgroup$ – Aaron Stevens Nov 11 '18 at 5:15
  • $\begingroup$ @AaronStevens Yes. $\endgroup$ – Dannyu NDos Nov 11 '18 at 5:35
  • $\begingroup$ In a uniformly polarized object, $\rho_b=0$ $\endgroup$ – Aaron Stevens Nov 11 '18 at 5:39
  • $\begingroup$ @AaronStevens And $\nabla \cdot \vec{P} = 0$. This is a trivial example and not quite interesting... $\endgroup$ – Dannyu NDos Nov 11 '18 at 5:41
  • $\begingroup$ Well you never said that example wasn't allowed :) $\endgroup$ – Aaron Stevens Nov 11 '18 at 5:42
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This is a temporary answer from self-education. You can verify or edit this answer. – Dannyu NDos

Intuitively, bound charge ($\rho_b$) can be thought as a charge caught in a specific point of an object. For example, atomic nuclei. On the contrary, free charge ($\rho_f$) is a charge that can move freely around. For example, free electrons, or amorphous ions. In this sense, it is even more obvious that bound charge is conserved.

$\rho_b$ can change in time if the object itself restricting the bound charges moves. In this case, we introduce the concept of polarization ($\vec{P}$) and have the equation $\frac{\partial \rho_b}{\partial t} = - \frac{\partial}{\partial t}(\nabla \cdot \vec{P})$. (This is a "demonstration" of the desired situation.)

In the dielectric of a capacitor, since the dielectric always has zero total charge inside, $\rho_b = 0$ anytime and everywhere. There's only free charges that builds up in both plates. So "In a capacitor, free charges turn bound, or vice versa" is a misconception. It should be "In a capacitor, free current turns into displacement currect, or vice versa."

Free charges can turn bound or vice versa if the material of the object changes. For example, if the object gets an electrolysis, or if the object turns into plasma. In this case, still, total charge is conserved.

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