Bound charges in the surface of a dielectric

Question

I am trying to understand how $\sigma_b=\vec P \cdot \hat n$.

In wikipedia I found the correct derivation but I don't understand some things. There it is said:

The bound surface charge is the charge piled up at the surface of the time dielectric, given by the dipole moment perpendicular to the surface: $q_b=\frac {\vec d \cdot \hat n}{|\vec s|}$ where $\vec s$ is the separation between the point charges constituting the dipole, $\vec d$ is the electric dipole moment, $\hat n$ is the unit normal vector to the surface.

Where does this come from? That bound charge is related somehow to the electric dipole. Plus the electric dipole has zero net charge, if you would pile them up in the surface wouldn't you get zero charge and zero surface charge density as a result?

Then we have:

Taking infinitesimals: $dq_b=\frac{d \vec d}{|\vec s|}\cdot \hat n$

and dividing by the differential surface element dS gives the bound surface charge density:

$$\sigma_b=\frac{dq_b}{dS}=\frac{d \vec d}{|\vec s|dS} \cdot \hat n=\frac{d \vec d}{dV} \cdot \hat n$$

Why is : $|\vec s|dS=dV$ ?

Answer 1

The wiki page needed a small correction on the distance $|\vec s|$. It should be the vertical distance $s_\perp$. Also, the notations were indeed greatly causing confusion.

Let me use the following graphic to explain. A dipole moment is shown as the red arrow, from $-q$ to $+q$ with separation $\vec s$, the moment defined as $$ \vec p = q\, \vec s. $$

And the normal unit vector $\hat n$ point on the surface outward. The surface charge induced by this dipole moment is $+q$: $$ q = \frac{|\vec p|}{|\vec s|} = \frac{\vec p \cdot \hat n}{|\vec s|_\perp} $$ where the $|\vec s|$ is replaced by $|\vec s|_\perp$ due to the inner product between $\vec p$ and $\hat n$.

Then, the surface charge density is the ratio between the small number of charges, $\Delta q$, and the small area $\Delta S$: \begin{align} \sigma_b & = \lim_{\Delta S \to 0} \frac{\Delta q}{\Delta S}\\ &= \lim_{\Delta S \to 0} \frac{\Delta \vec p \cdot \hat n}{ |\vec s|_\perp \Delta S}\\ &= \lim_{\Delta S \to 0} \frac{\Delta \vec p \cdot \hat n}{ \Delta V}\\ &= \vec P \cdot \hat n \end{align} The small volume $\Delta V$ is the region near the surface $\Delta S$ with length $|\vec s|_\perp$ into the bulk, and the polarization $\vec P$ defined as the ratio between number of dipoles $\Delta \vec p$ within the small volume $\Delta V$: $$\vec P = \lim_{\Delta V \to 0} \frac{\Delta \vec p}{\Delta V} $$.