25 Jun 2025

Electric Fields in Dielectrics

Polarization and polarizability, bound charge, susceptibility, dielectric constant, displacement field, dielectric Gauss law, and Clausius-Mossotti theory.

electricity-and-magnetism dielectrics polarization displacement-field clausius-mossotti

An applied electric field slightly separates positive and negative charge inside matter and can also orient permanent molecular dipoles. The macroscopic electric dipole moment per unit volume is the polarization

\[\boxed{\mathbf P(\mathbf r) =\frac{\text{electric dipole moment}}{\text{volume}}},\]

with SI unit $\mathrm{C\,m^{-2}}$.

Polarization and bound charge

A volume element $\mathrm d\tau’$ has dipole moment $\mathbf P(\mathbf r’)\,\mathrm d\tau’$. Its potential at $\mathbf r$ is

\[\mathrm dV =\frac{1}{4\pi\epsilon_0} \mathbf P(\mathbf r')\cdot \boldsymbol\nabla'\!\left(\frac{1}{\lvert\mathbf r-\mathbf r'\rvert}\right) \mathrm d\tau'.\]

Use

\[\boldsymbol\nabla'\cdot \left(\frac{\mathbf P}{\lvert\mathbf r-\mathbf r'\rvert}\right) =\frac{\boldsymbol\nabla'\cdot\mathbf P}{\lvert\mathbf r-\mathbf r'\rvert} +\mathbf P\cdot\boldsymbol\nabla' \left(\frac{1}{\lvert\mathbf r-\mathbf r'\rvert}\right).\]

Integration over the polarized body and the divergence theorem give

\[V(\mathbf r)=\frac{1}{4\pi\epsilon_0} \left[ \oint_S\frac{\mathbf P\cdot\hat{\mathbf n}} {\lvert\mathbf r-\mathbf r'\rvert}\,\mathrm da' +\int_V\frac{-\boldsymbol\nabla'\cdot\mathbf P} {\lvert\mathbf r-\mathbf r'\rvert}\,\mathrm d\tau' \right].\]

The field of polarized matter is therefore the field of the equivalent bound charges

\[\boxed{\rho_b=-\boldsymbol\nabla\cdot\mathbf P}, \qquad \boxed{\sigma_b=\mathbf P\cdot\hat{\mathbf n}}.\]

Uniform $\mathbf P$ gives $\rho_b=0$ inside, but generally leaves bound charge on surfaces whose normal has a component along $\mathbf P$.

A uniform slab carries opposite bound surface charges. In an isotropic dielectric, the Lorentz cavity contributes $\mathbf P/(3\epsilon_0)$ to the molecular local field.

Electric field and displacement field in matter

The total charge is $\rho=\rho_f+\rho_b$, where $\rho_f$ denotes charge not included in the polarization description. Gauss’s law is

\[\boldsymbol\nabla\cdot\mathbf E =\frac{\rho_f-\boldsymbol\nabla\cdot\mathbf P}{\epsilon_0}.\]

Move the polarization term to the left and define

\[\boxed{\mathbf D=\epsilon_0\mathbf E+\mathbf P}.\]

Then Gauss’s law in a dielectric becomes

\[\boxed{\boldsymbol\nabla\cdot\mathbf D=\rho_f}, \qquad \boxed{\oint_S\mathbf D\cdot\mathrm d\mathbf a=Q_{f,\mathrm{enc}}}.\]

For an interface with unit normal from medium 1 to medium 2, a pillbox gives

\[\boxed{\hat{\mathbf n}\cdot(\mathbf D_2-\mathbf D_1)=\sigma_f}.\]

Electrostatics still has $\boldsymbol\nabla\times\mathbf E=0$, so a narrow loop gives

\[\boxed{\hat{\mathbf n}\times(\mathbf E_2-\mathbf E_1)=0}.\]

Only the free surface charge appears in the normal-$\mathbf D$ condition; bound charge is already contained in $\mathbf P$.

Susceptibility, dielectric constant, and polarizability

For a linear, isotropic dielectric,

\[\boxed{\mathbf P=\epsilon_0\chi_e\mathbf E},\]

where the electric susceptibility $\chi_e$ is dimensionless. Hence

\[\mathbf D =\epsilon_0(1+\chi_e)\mathbf E =\epsilon\mathbf E =\epsilon_0\epsilon_r\mathbf E,\]

so

\[\boxed{\epsilon_r=1+\chi_e}, \qquad \boxed{\epsilon=\epsilon_0\epsilon_r}.\]

The relative permittivity $\epsilon_r$ is also called the dielectric constant in the static, linear regime.

Microscopically, an isotropic molecule with induced dipole moment $\mathbf p$ has polarizability $\alpha$ defined by

\[\boxed{\mathbf p=\alpha\mathbf E_{\mathrm{loc}}}.\]

In SI, $[\alpha]=\mathrm{C\,m^2\,V^{-1}}=\mathrm{F\,m^2}$. The local field $\mathbf E_{\mathrm{loc}}$ acting on a molecule need not equal the macroscopic field $\mathbf E$.

Clausius-Mossotti equation

For a homogeneous isotropic or cubic dielectric, the Lorentz spherical-cavity construction gives

\[\boxed{\mathbf E_{\mathrm{loc}} =\mathbf E+\frac{\mathbf P}{3\epsilon_0}}.\]

If $N$ is the molecular number density, then

\[\mathbf P=N\mathbf p =N\alpha\left(\mathbf E+\frac{\mathbf P}{3\epsilon_0}\right).\]

Collect the $\mathbf P$ terms:

\[\mathbf P\left(1-\frac{N\alpha}{3\epsilon_0}\right) =N\alpha\mathbf E.\]

Insert $\mathbf P=\epsilon_0(\epsilon_r-1)\mathbf E$, cancel $\mathbf E$, and define

\[x=\frac{N\alpha}{3\epsilon_0}.\]

Then

\[(\epsilon_r-1)(1-x)=3x.\]

Expanding and collecting $x$,

\[\epsilon_r-1=x(\epsilon_r+2).\]

Therefore

\[\boxed{\frac{\epsilon_r-1}{\epsilon_r+2} =\frac{N\alpha}{3\epsilon_0}}.\]

The derivation assumes a linear, homogeneous, isotropic or cubic, nonpolar dielectric whose molecules can be treated as weakly interacting polarizable units. Strong correlations, anisotropy, permanent-dipole orientation, or large fields require a more detailed model.

The bound-charge identity, interface relations, and Clausius-Mossotti algebra are verified with exact zero residuals in the Unit II dielectric worksheet.