📂Electrodynamics

Electric Field Created by Bound Charges and Polarized Objects

Bound Charges

External electric fields cause the dipoles in a material to align in one direction, polarizing the material and giving it a dipole moment $\mathbf{p}$. The electric field produced by these dipole moments is calculated as follows. The potential created by the dipole moment $\mathbf{p}$ is as follows.

$$\begin{equation} V(\mathbf{r}) = \dfrac{1}{4 \pi \epsilon_{0}} \dfrac{ \mathbf{p} \cdot \crH } {\cR ^2} \label{1} \end{equation}$$

$\mathbf{r}^{\prime}$ is the position vector of the source point, $\mathbf{r}$ is the position vector of the observation point, $\bcR=\mathbf{r} - \mathbf{r}^{\prime}$ is the separation vector. The polarization density $\mathbf{P}$ is the dipole moment per unit volume, so

$$\mathbf{P}=\dfrac{\mathbf{p}}{d \tau^{\prime}}$$

Substituting this into $\eqref{1}$,

$$V(\mathbf{r}) = \dfrac{1}{4 \pi \epsilon_{0} } \displaystyle \int _\mathcal{V} \dfrac{ \mathbf{P (\mathbf{r}^{\prime} ) \cdot \hat { \boldsymbol {\cR} } } }{\cR ^2} d \tau^{\prime}$$

Using $\nabla ^{\prime} \left( \dfrac{1}{\cR} \right) = \dfrac{ \crH } {\cR^2}$ here, the above equation becomes

$$V=\dfrac{1}{4\pi \epsilon_{0}} \displaystyle \int _\mathcal{V} \mathbf{P} \cdot \nabla^{\prime} \left( \dfrac{1}{\cR} \right) d \tau^{\prime}$$

Product rule involving the del operator

$$\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$$

Using the above multiplication rule,

$$\begin{align*} && \nabla^{\prime} \cdot \left( \dfrac{ \mathbf{P} } {\cR} \right) =&\ \dfrac{1}{\cR} ( \nabla^{\prime} \cdot \mathbf{P} ) + \mathbf{P} \cdot \nabla^{\prime} \left( \dfrac{1}{\cR} \right) \\ \implies && \mathbf{P} \cdot \nabla^{\prime} \left( \dfrac{1}{\cR} \right) =&\ \nabla^{\prime} \cdot \left( \dfrac{ \mathbf{P} } {\cR} \right) -\dfrac{1}{\cR} ( \nabla^{\prime} \cdot \mathbf{P} ) \end{align*}$$

and substituting this,

$$V = \dfrac{1}{4 \pi \epsilon_{0}} \int_\mathcal{V} \nabla^{\prime} \cdot \left( \dfrac{ \mathbf{ P} }{\cR} \right) d\tau^{\prime} -\dfrac{1}{4 \pi \epsilon_{0}} \int _\mathcal{V} \dfrac{1}{\cR} (\nabla^{\prime} \cdot \mathbf{P} ) d\tau^{\prime}$$

Divergence theorem

$$\int_\mathcal{V} \nabla \cdot \mathbf{ F} dV = \oint _\mathcal{S} \mathbf{F} \cdot d \mathbf{ S}$$

Using the divergence theorem on the first term,

$$V = \dfrac{1}{4 \pi \epsilon_{0}} \oint_\mathcal{S} \left( \dfrac{ \mathbf{ P} }{\cR} \right) \cdot d\mathbf{a}^{\prime} -\dfrac{1}{4 \pi \epsilon_{0}} \int _\mathcal{V} \dfrac{1}{\cR} (\nabla^{\prime} \cdot \mathbf{P} ) d\tau^{\prime}$$

If the unit normal vector to the surface is denoted by $\hat { \mathbf{n} }$, it can be represented as $\mathbf{P} \cdot d\mathbf{a}^{\prime}=\mathbf{P} \cdot \hat{ \mathbf{n} } da^{\prime}$. Here, $\mathbf{P} \cdot \hat{ \mathbf{n} }$ is denoted as $\sigma_{b}$ and called the bound surface charge density.

$$\sigma_{b} = \mathbf{P} \cdot \hat{ \mathbf{n} }$$

Similarly, $-\nabla^{\prime} \cdot \mathbf{P}$ is denoted as $\rho_{b}$ and called the bound volume charge density.

$$\rho_{b}=-\nabla^{\prime} \cdot \mathbf{P}$$

Now, the potential due to the polarization density $\mathbf{P}$ can be expressed as the potential produced by these two bound charges.

$$V(\mathbf{r}) = \dfrac{1}{4 \pi \epsilon_{0}} \oint_\mathcal{S} \dfrac{ \sigma_{b}} {\cR} da^{\prime}+\dfrac{1}{4 \pi \epsilon_{0}} \int_\mathcal{V} \dfrac{\rho_{b}}{\cR} d\tau^{\prime}$$

The potential created by a polarized body is the sum of the potentials created by the bound volume charge density $\rho_{b}$ and the bound surface charge density $\sigma_{b}$.

Characteristics

1. The total bound charge adds up to $0$. Polarizing an electrically neutral dielectric moves charges to create bound charges, but the total charge still equals $0$.

2. When the polarization density is uniform, the bound volume charge density equals $0$. Since $\rho_{b}=-\nabla \cdot \mathbf{P}$, if $\mathbf{P}$ is constant, the differentiated result will be $0$.