Bound Charges

By definition, polarization is defined as dipole moment density $$\textbf{p} = \int \textbf{P} d\tau$$

Bound Charges

Potential due to a dipole is $$V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\hat{d}\cdot\textbf{p}}{d^2}$$ where $\textbf{d} = \textbf{r} - \textbf{r}'$. If the dipole is expressed in polarization, the potential can be written as $$\begin{align} V(\textbf{r}) &= \frac{1}{4\pi\epsilon_0}\int \frac{\hat{d}\cdot \textbf{P}(\textbf{r}')}{d^2}d\tau \\ &= \frac{1}{4\pi\epsilon}\int \textbf{P}\cdot\nabla(\frac{1}{d})d\tau \\ &= \frac{1}{4\pi\epsilon}\int(\nabla\cdot(\frac{\textbf{P}}{d}) - \frac{1}{d}(\nabla\cdot\textbf{P}))d\tau \\ &= \frac{1}{4\pi\epsilon}\oint\frac{1}{d}\textbf{P}\cdot d\textbf{a} - \frac{1}{4\pi\epsilon}\int\frac{1}{d}(\nabla\cdot\textbf{P})d\tau \end{align}$$ This resembles the expression of electric potential by charges. We identify the elements that generate this kind of "electric potential". Define bound surface charge to be $$\bbox[5px,border:2px solid #666] { \sigma_b = \textbf{P}\cdot\hat{\textbf{n}} }$$ and bound volume charge to be $$\bbox[5px,border:2px solid #666] { \rho_b = \nabla\cdot\textbf{P} }$$ The potential can be written as $$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\oint\frac{\sigma_b}{d}da + \frac{1}{4\pi\epsilon_0}\int\frac{\rho_b}{d}d\tau$$ The total charges due to these bound charges overall is always zero $$\begin{align} Q_{tot} &= \oint\sigma_bda + \int\rho_bd\tau \\ &= \oint\textbf{P}\cdot d\textbf{a} - \int\nabla\cdot\textbf{P}d\tau \\ &= \int\nabla\cdot\textbf{P}d\tau - \int\nabla\cdot\textbf{P}d\tau \\ &= 0. \end{align}$$ These charges are called bound charges because they are "bound", ie not free to move (or else they would have flowed and neutralize each other). It is the special arrangement of the charges that give rise to the electric potential. There exists non-zero electric potential despite the zero net charges.

Example (Griffiths Third Edition Ex 4.2)

Find the electric field produced by a uniformly polarized sphere of radius $R$

Solution: Choose a coordinate system with $z$ parallel to the direction of polarization and origin at the sphere centre. $\rho_b = 0$ and $\sigma_b= \vec{P}\cdot\hat{n}=P\cos\theta$. Using separation of variables in radial coordinate, by the previous example, we know that the potential of a spherical shell of radius $R$ with charge density $\sigma_0(\theta$ is $$V(r,\theta) = \begin{cases} \sum^{\infty}_{l=0}A_lr^lP_l(\cos\theta), \text{ if } r < R\\ \sum^{\infty}_{l=0}\frac{A_lR^{2l+1}}{r^{l+1}}P_l(\cos\theta), \text{ if } r \geq R\\ \end{cases}$$ where $$\sum^{\infty}_{l=0}(2l+1)A_lR^{l-1}P_l(\cos\theta)=\frac{1}{\epsilon_0}\sigma_0(\theta)$$ Substitue $\sigma_0(\theta)=\sigma_b=P\cos\theta$, $P_l$ terms other than $l=1$ are all 0. $$(3)A_1R^{0}=\frac{1}{\epsilon_0} \implies A_1 = \frac{P}{3\epsilon_0}$$ So, the potential is $$V(r,\theta) = \begin{cases} \frac{P}{3\epsilon_0}r\cos\theta, \text{ if } r < R\\ \frac{P}{3\epsilon_0}\frac{R^{3}}{r^{2}}\cos\theta, \text{ if } r \geq R\\ \end{cases}$$ for $r < R$, $$\vec{E}=-\nabla V = -\frac{P}{3\epsilon_0}\hat{z}=-\frac{1}{3\epsilon_0}\vec{P}$$ for $r \geq R$, the potential is $$V = \frac{P}{3\epsilon_0}\frac{R^{3}}{r^{2}}\cos\theta = \left(\frac{4}{3}\pi PR^3\right)\frac{\cos\theta}{4\pi\epsilon_0r^2}$$ which is equivalent to a perfect dipole with dipole moment $$\vec{p} = \frac{4}{3}\pi R^3\vec{P}$$ Thus, the field is $$\vec{E}(r,\theta)=\frac{PR^3}{3\epsilon_0 r^3}(2\cos\theta\hat{r}+\sin\theta\hat{\theta})$$

Checkpoint (Griffiths Third Edition Q4.10)

A sphere of radius $R$ carries a polarication $$\vec{P}(\vec{r})=k\vec{r},$$ where $k$ is a constant and $\vec{r}$ is the vector from the centre. Calculate the bound charges $\sigma_b$ and $\rho_b$. Find the field inside and outside the sphere.

$$\sigma_b = \vec{P}\cdot\hat{n}=kR$$ $$\begin{align} \rho_b &= -\nabla \cdot\vec{P}\\ &= -\frac{1}{r^3}\frac{\partial}{\partial r}(r^2kr)\\ &= -\frac{1}{r^2}3kr^2\\ &= -3k \end{align}$$ Using Gauss's law, for $r<R$, $$\begin{align} \oint\vec{E}\cdot d\vec{a}&=\frac{1}{\epsilon_0}Q_{enc}\\ E(4\pi r^2) &= \frac{1}{\epsilon_0}\frac{4}{3}\pi r^3\rho_b\\ \vec{E} &= \frac{1}{3\epsilon_0}\rho_b r\hat{r}\\ &=-\frac{k}{\epsilon_0}\vec{r} \end{align}$$ Using Gauss's law, for $r>R$, $$\begin{align} \oint\vec{E}\cdot d\vec{a}&=\frac{1}{\epsilon_0}Q_{enc}\\ E(4\pi r^2) &= \frac{1}{\epsilon_0}\left(\frac{4}{3}\pi R^3\rho_b + (\sigma_b)(4\pi R^2)\right)\\ &= \frac{1}{\epsilon_0}\left(\frac{4}{3}\pi R^3(-3k) + (kR)(4\pi R^2)\right) =0\\ \vec{E} &= 0 \end{align}$$

Electric Displacement

By Gauss's Law, $$\epsilon_0\nabla\cdot\textbf{E} = \rho = \rho_b + \rho_f$$ where $\rho_b$ is the charge density due to bound charges while $\rho_f$ is the charge density due to free charges. In obtaining the Gauss's law, we use the charge density (a function of displacement). The charge density does not specify whether the charges are free to flow or bound. Thus, $\textbf{E}$ here is the E-field due to both free charges and bound charges, i.e. polarization. Express in terms of polarization, we have $$\begin{align} \epsilon_0\nabla\cdot\textbf{E} &= \rho_b + \rho_f \\ &= -\nabla\cdot\textbf{P} + \rho_f \\ \nabla\cdot(\epsilon_0\textbf{E} + \textbf{P}) &= \rho_f. \end{align}$$ Define the electric displacement as $$\bbox[5px,border:2px solid #666] { \textbf{D} = \epsilon_0\textbf{E} + \textbf{P} }$$ The electric field includes contributions by both the free charges and bound charges, hence can be undermined by the existence of bound charges. The electric displacement represents the electric field solely contributed by the free charges.
With the definition of the electric displacement, the Gauss's law can be written as $$\bbox[5px,border:2px solid #666] { \nabla\cdot\textbf{D} = \rho_f }$$ or $$\bbox[5px,border:2px solid #666] { \oint\textbf{D}\cdot d\textbf{a} = Q_{f_{enc}} }$$ Note that Gauss's law is not modified, but just express in terms of electric displacement. By Gauss's law, we know that $$D^{\bot}_{above} - D^{\bot}_{below} = \sigma_f$$ The curl of electric displacement is $$\nabla\times\textbf{D} = \epsilon_0(\nabla\times\textbf{E}) + (\nabla\times\textbf{P}) = \nabla \times \textbf{P}$$ as curl of polarization is not necessarily zero, curl of electric displacement is not necessarily zero, i.e. one cannot define potential for electric displacement. Since electric displacement is not curl free, unlike electric field, its parallel components boundary condition is $$D^{\parallel}_{above} - D^{\parallel}_{below} = P^{\parallel}_{above} - P^{\parallel}_{below}$$

Example (Griffiths Third Edition Ex 4.4)

A long straight wire, carrying uniform line charge $\lambda$, is surrounded by rubber insulation out to a radius $a$. Find the electric field outside the wire.

Solution: Draw a cylindrical Gaussian surface with radius $s$ and length $L$ $$\begin{align} \oint \vec{D}\cdot d\vec{a} &= Q_{f_{enc}}\\ D(2\pi sL) &= \lambda L\\ \vec{D} &= \frac{\lambda}{2\pi s}\hat{s} \end{align}$$ Outside the rubber, $s>a$, $\vec{P}=0$ $$\vec{E}=\frac{1}{\epsilon_0}\vec{D}=\frac{\lambda}{2\pi\epsilon_0 s}\hat{s}.$$

Checkpoint (Griffiths Third Edition Q4.15)

A thick spherical shell (inner radius $a$, outer radius $b$) is made of dielectric material with a "frozen-in" polarization $$\vec{P}(\vec{r})=\frac{k}{r}\hat{r},$$ where $k$ is a constant and $r$ is the distance from the center. (There is no free charge in the problem.) Find the electric field in all three regions.

$$\begin{align} \oint \vec{D}\cdot d\vec{a} &= Q_{f_{enc}} = 0\\ \vec{D} &= 0\\ \vec{D} &= \epsilon_0\vec{E}+\vec{P} = 0\\ \vec{E} &= -\frac{1}{\epsilon_0}\vec{P}. \end{align}$$ For $r< a$ and $r > b$, $$\vec{E}=0.$$ For $a < r < b$, $$\vec{E} = -\frac{k}{\epsilon_0r}\hat{r}$$