# Bound charges in the surface of a dielectric

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

## 1 Answer

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 $$\Delta V$$: $$\vec P = \lim_{\Delta V \to 0} \frac{\Delta \vec p}{\Delta V}$$.

• Why are you considering the case where the positive charge of the dipole is close to the surface, can't we have both charges of the dipole in the surface, in other words the direction of the dipole moment is parallel to the surface? I still don't understand how you came up with that perpendicular $|\vec s|$. by length $|\vec s|_{perpendicular}$ you mean, the volume is the result of the surface $\Delta S$ and "depth" $|\vec s|_{perpendicular}$ from the surface of the material to a point inside of it? Feb 1, 2022 at 21:43
• Induced bound surface charge is arising from the positive (or negative) remnant on the surface. You cannot have both type of charges on the surface, It cannot for polarization in that way $\vec P = 0$.
– ytlu
Feb 1, 2022 at 21:50
• Ok. I understand your setup. Basically hom. polarisation, which generates poz/negativ charges in the opposite surfaces. But still I don't get the other two things I mentioned. I understand how you get the normal component of S, but I don't understand why you are doing the inner product with the surface normal. + $|\vec p|$ doesn't imply $\vec p \vec n$ Feb 1, 2022 at 21:51
• Because the direction of dipole is not necessary perpendicular to the surface, $\vec P$ is not parallel to $\hat n$. In order to calculate the volume occupied by a dipole layer, the volume is surface area times the vertical distance.
– ytlu
Feb 1, 2022 at 21:55
• And the reason why you want to calculate the volume occupied by a dipole layer, is because this, indirectly, tells you the amount of charge that is being displacement in the surface? If what I said is correct, then the smaller the angle between the dipole moment and the normal, the bigger the charge value, because more dipoles, aligned parallel with each other and perpendicular to the surface, bring more charges to it? Do I make sense? Feb 1, 2022 at 22:00