Why is it that the bound charge is $Q_b = – \oint_S{\mathbf{P} \cdot d\mathbf{S}}$? In particular, why is there a negative sign? Hayt's book on electromagnetism describes this as the "net increase in bound charge within the closed surface". Compared to Gauss' law $Q_{f} = \oint_S{\mathbf{D} \cdot d\mathbf{S}}$, the free charge is just the electric flux, so I don't see why the bound charge has a negative sign.
Also, if the electric flux is independent of bound charge, how come the D-field changes across material boundaries? I thought the D-field was defined as $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$ to cancel out the effect of the bound charge (since the effect of polarization is already incorporated into the E-field). Is the D-field dependent on bound charge?
Just to add to the second part of my question. I know that for a dielectric-dielectric boundary the tangential components $\frac{D_{t1}}{\epsilon_1} = \frac{D_{t2}}{\epsilon_2}$ and if $Q_{free} = 0$ on the boundary, then the normal components $D_{n1} = D_{n2}$. The tangential component will not contribute to the flux integral so for any $\epsilon_1$ and $\epsilon_2$ across a boundary the electric flux integral will evaluate to 0. It's clear that the flux is independent of the bound charge, but the the tangential component changes so the D-field is not?
Best Answer
Polarization $\mathbf P$ is average electric moment per unit volume. Electric moment of a neutral group of charges is a vector that points from the center of negative charge to the center of positive charge. When some closed surface $S$ with outward normal $\mathbf n$ is chosen inside a medium, expected amount of charge that have been transported from the interior to the exterior of the surface through element $dS$ upon polarization is given by $\mathbf P \cdot \mathbf n dS$. Since this charge is outside, the net charge inside is of equal magnitude and opposite sign, hence the charge inside is
$$ Q_b = -\oint_S \mathbf P\cdot d\mathbf S. $$