# Total bound charge equal to zero (proof)

I always see the assumption that the total bound charge is always zero, but is it not clear for me why is it true for every dielectric material, and I haven't find it in a textbook.

I know that $$\sigma _b =\overrightarrow{P}\cdot \widehat{n}$$ and $$\rho _b =- \nabla\cdot \overrightarrow{P}$$, and I see that this must be true, $$\oint_{S}^{}\overrightarrow{P}\cdot \widehat{n}\,ds - \int_{V}^{}\nabla\cdot \overrightarrow{P}dV =0,$$ but I don't know how to proceed from here.

• In some ways it's better to assume that the total bound charge on a dielectric medium is zero, and then find a $\vec{P}$ for which $\nabla \cdot \vec{P} = \rho_b$. See my answer here for a description of this perspective. Nov 6, 2023 at 12:57

The total bound charge on a body $$\Omega$$ (with boundary $$\partial\Omega$$) is obviously the sum of the bound surface and bound volume charges, $$Q_{b}=\int_{\Omega} dV\,\rho_{b}+\int_{\partial\Omega}dA\,\sigma_{b}=-\int_{\Omega} dV\left(\vec{\nabla}\cdot\vec{P}\right)+\int_{\partial\Omega}dA\left(\vec{P}\cdot\vec{n}\right).$$ This is what you have concluded has to be vanishing. However, that this is zero is actually just Gauss's theorem (the divergence theorem), $$\int_{\Omega} dV\left(\vec{\nabla}\cdot\vec{P}\right)=\int_{\partial\Omega}d\vec{S}\cdot\vec{P},$$ since $$d\vec{S}=\hat{n}\,dA$$.
In a conductor the electrons are nearly free to move in the surface and thus constituing electric current while in dielectric materials the electrons are bound in molecules and at most the molecules or atoms constitute local dipoles that can rotate or being deformate showing partial charge density. Polarization can be due to change in orientation of the molecule or to the deformation of the electric "cloud" when an external electric field is applied. Bound charges set up electric dipoles in response to an applied electric field $$E$$, and polarize other nearby dipoles tending to line them up, the net accumulation of charge from the orientation of the dipoles is the bound charge.
Considering the Gauss theorem $$\oint_S \vec{E} \cdot d\vec{s}=\frac{Q_{enclosed}}{\epsilon_0}$$ or equivalently $$\oint_S \vec{D} \cdot d\vec{s}=Q_{enclosed}$$ It says that the total flux through a closed surface (Gauss surface), boudary of a volume $$S=\partial V$$ is equal to the algebraic sum of the charges enclosed inside it. Thus, because a molecule is neutral (the sum of positive and negative charges is always zero) the bound charge inside the gauss sphere doesn't count and you can only see the effects of the net charge enclosed (free charges)