# Dielectric in a parallel plate capacitor

Uniform charge: each atom has charge $q$. Magnitude of dipole moment is $q s$, where $s$ is the distance the nucleus is shifted. According to my notes, the charge on the surface of a dielectric in between the plates is $N q s S$, where $N$ is the number of dipoles and $S$ is the surface area of the plate.

But surely this should be $N q s$, because the charge should on the surface should be the same irrespective of the surface area (because we are using the number of charges on the surface $N$).

• The units still wouldn't be right. Maybe N is number density (number per unit volume) instead. I think the original formula works better then. – Art Brown May 14 '12 at 0:03
• yes I miagine the original formula works (because it helps derive the relative permitivity ...), but qualitatively, why does my reasoning fail? – Adam Rubinson May 14 '12 at 1:11
• I think your reasoning is sound, provided you write the net charge as $(N_l/\delta)qs$, where $N_l$ is the number of charges in the surface layer (imagining the dielectric as a stack of layers), $\delta$ is the distance between layers, and $q$ and $s$ are as before. [$N_l/\delta=NS$, with $N$ the number density]. – Art Brown May 14 '12 at 1:59

## 1 Answer

There are two misconceptions present in your explanation of the problem.

• $N$ is not number of dipoles, but their volumetric density
• $Q$ is not total charge, but equivalent charge at boundaries of the dielectric.

The idea is that (a) dielectric of the area $A$ and height $L$ polarized homogeneously along its height and (b) two plan-parallel plates of the area $A$, distanced by $L$ and with charges $N A p$ and $-N A p$ produce macroscopically the same electric field ($N$ is volumetric density and $p = q s$ is polarization of one dipole).

This effect can be relatively simply understood if you imagine that you have charges of volumetric density $N$ homogeneously distributed all along material. Initially positive and negative charges are on the same positions, all material is electrically neutral and polarization equals zero (picture left). Now you pull all positive charges for $s/2$ up and all negative charges for $s/2$ down (picture right), so you actually get total dipole moment $P' = N V p = N A L q s$. Figure: red = positive charge, blue = negative charge, violet = neutral.

What is the effect of such movements? The bulk of dielectric material remains neutral in terms of charge, but you do get excess charge $Q = N A s q$ at the top and excess charge $-Q = -N A s q$ at the bottom of the dielectric ($A s$ is the volume at the top or bottom where only one type of charge is present).

The point of this simple derivation is that surface charge density $\sigma = \frac{Q}{A} = N s q$ equals polarization volumetric density $P = \frac{P'}{A L} = N q s$, i.e. $\sigma = P$. (Polarization density is by definition total dipole moment of the dielectric divided by its volume.)

• Pygmalion, my misunderstanding was not that huge actually, and I know what is "going on" (what you have illustrated). I just thought N was number of charges on the plates, although actually it is charge density. The rest is obvious (to me). Maybe I'm missing something... – Adam Rubinson May 14 '12 at 19:06
• @AdamRubinson So I guess you understand now? $N$ is volumetric density of charges, while $Q$ is total surface charge. – Pygmalion May 14 '12 at 19:14
• Edit: Yes, and q is atom charge. So you are looking at the big picutre here, not just at the surfaces (like I WAS). Edit: I was using the same definition of Q as you (my "q" is your "q"). On closer inspection, before I didn't use that As is the volume we are interested in. Your derivation is easy and nice. Fair play – Adam Rubinson May 14 '12 at 19:17
• But in your last line, shouldn't sigma = surface charge density? – Adam Rubinson May 14 '12 at 19:25
• @AdamRubinson Yep, right, I'll correct :) – Pygmalion May 14 '12 at 19:26