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Yet, $\mathbf J=\rho_f \mathbf V$ (where $\rho_f $ is the free charge density), and since $\rho_f=0$ , $\mathbf J$ and $\mathbf E$ should be zero The problem that you are running into is that this statement is not really correct. It should actually be: $\mathbf J_f= \Sigma \rho_i \mathbf V_i$ (where $\rho_i$ is the charge density of the i’th type of free ...


2

It is also the electric force, but caused by atoms/molecules of the insulating solid. Such an electric force is also found in conductors; the difference between conductors and insulators is in its implications on the mobility of charge carriers. And this is fully governed by quantum theory of solids: in short, in insulators an electron can not get into ...


2

the charge on the ungrounded conductor produces an electrostatic field which either attracts or repels electrons in the grounded conductor. The ground can be thought of as a huge reservoir of electrons at zero potential which can furnish electrons or disperse them. So, if your conductor has a + charge, it will attract electrons in the grounded conductor, and ...


1

$\rho$ is zero inside a resistor, because the positive and negative charges cancel each other. $J$ is non-zero since only electrons do the moving.


1

You have mis-stated Gauss's Law. Gauss's Law can be written as either $${\bf\nabla}\cdot{\bf E}=\frac{\rho}{\epsilon_0}$$ where $\rho$ is the total charge (not the free charge); or as $${\bf\nabla}\cdot{\bf D}=\rho_f$$ where $\bf D$ is the electric displacement field and $\rho_f$ is the free charge. So we don't have 0 free charge (if we did, $\sigma$ would ...


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