= Solution
\b[From molecular dipoles to macroscopic fields.] A <dielectric> contains charges which remain bound to its atoms or <molecules>. An applied <electric field> displaces positive and negative charge centres and can orient permanent <electric dipoles>. The <electric polarization> $\mathbf P$ is <electric dipole moment> per unit <volume (thermodynamics)>, in $\mathrm{C\,m^{-2}}$. Its associated <bound charge> is
$$
\rho_b=-\nabla\cdot\mathbf P,\qquad \sigma_b=\mathbf P\cdot\widehat{\mathbf n}
$$
at the boundary of the polarized material, where $\widehat{\mathbf n}$ points outward. A uniform <electric polarization> has no <volume (thermodynamics)> <bound charge>, but generally has surface <bound charge>. The <electric field> remains the <force> per unit test charge; it includes the fields of both free and bound charges.
Introducing the <electric displacement field> $\mathbf D=\epsilon_0\mathbf E+\mathbf P$ packages the free-charge source separately:
$$
\nabla\cdot\mathbf E=\frac{\rho_f+\rho_b}{\epsilon_0},\qquad
\nabla\cdot\mathbf D=\rho_f,\qquad \nabla\times\mathbf E=0.
$$
For a homogeneous linear isotropic <dielectric>,
$$
\mathbf P=\epsilon_0\chi_e\mathbf E,\qquad
\mathbf D=\epsilon\mathbf E,\qquad \epsilon=\epsilon_0(1+\chi_e)=\epsilon_0\epsilon_r.
$$
The <electric susceptibility> $\chi_e$ is dimensionless; the <permittivity> $\epsilon$ has units $\mathrm{F\,m^{-1}}$. This is a constitutive approximation. An anisotropic <dielectric> requires a <permittivity tensor>, while a nonlinear <dielectric> need not have <electric polarization> proportional to <electric field>. The field in these equations is the macroscopic field, not necessarily the <local field> at an individual <molecule>.
\b[Interfaces.] Let $\widehat{\mathbf n}$ point from medium 1 into medium 2. Integrating <Gauss's law> over a thin pillbox, and the electrostatic <curl> equation around a thin loop, gives the <electrostatic interface boundary conditions>
$$
\boxed{\widehat{\mathbf n}\cdot(\mathbf D_2-\mathbf D_1)=\sigma_f,
\qquad \widehat{\mathbf n}\times(\mathbf E_2-\mathbf E_1)=0.}
$$
Thus normal electric <displacement> is continuous when there is no free surface charge, whereas tangential <electric field> is continuous even if the <permittivity> changes. Normal <electric field> generally jumps because the interface carries <bound charge>. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with <Laplace's equation> in each homogeneous charge-free region, determine the field.
\b[<Energy> and electrical sources.] Reversibly assembling <free charge> requires incremental <work> $\delta W=\int\mathbf E\cdot\delta\mathbf D\,dV$. For a linear <dielectric> whose response can be reversibly integrated from zero field, the <electrostatic energy> is
$$
\boxed{W=\frac12\int\mathbf E\cdot\mathbf D\,dV.}
$$
This includes the material polarization response, rather than simply using $\epsilon_0E^2/2$ everywhere. In a <capacitor>, $W=Q^2/(2C)=CV^2/2$. Inserting a <dielectric> increases <capacitance>. At fixed charge the stored <energy> decreases. At fixed voltage the stored field <energy> increases, but the voltage source supplies $V\,dQ$; the appropriate mechanical <energy> including that source changes by $-V^2dC/2$. Both cases predict attraction of <dielectric> into the <capacitor>. A <force> calculation must state which electrical quantity is fixed. A lossy or hysteretic <dielectric> also dissipates <energy>, so the reversible quadratic expression does not describe its full history.
\b[Shape and depolarization.] Suppose the <dielectric> is surrounded by <vacuum> and the distant imposed <electric field> is $\mathbf E_0$. Bound surface charge produces a <depolarization field>. For a uniformly polarized <ellipsoid>, along a <principal axis>,
$$
E_{\rm in}=E_0-\frac{N_dP}{\epsilon_0},\qquad
P=\epsilon_0(\epsilon_r-1)E_{\rm in},
\qquad
\boxed{E_{\rm in}=\frac{E_0}{1+N_d(\epsilon_r-1)}.}
$$
Here $N_d$ is the <depolarization factor of a dielectric ellipsoid>. This self-consistent field explains why the same material responds differently in different shapes.
An infinite slab with <electric field> normal to its faces has $N_d=1$, giving $E_{\rm in}=E_0/\epsilon_r$; its opposite surface bound charges reduce the internal field. For <electric field> parallel to the faces, $N_d=0$, giving $E_{\rm in}=E_0$. End effects are neglected.
A very long circular rod in an axial <electric field> has negligible depolarization away from its ends, so $E_{\rm in}\simeq E_0$. In a transverse <electric field> the infinite-cylinder solution has $N_d=1/2$, giving $E_{\rm in}=2E_0/(\epsilon_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written $\phi_{\rm in}=-B r\cos\theta$ and $\phi_{\rm out}=-E_0r\cos\theta+A\cos\theta/r$. Continuity of potential and normal electric <displacement> at $r=a$ gives $B=2E_0/(\epsilon_r+1)$.
For a <sphere> of radius $a$, $N_d=1/3$ and
$$
\boxed{\mathbf E_{\rm in}=\frac{3}{\epsilon_r+2}\mathbf E_0,
\qquad \mathbf p=4\pi\epsilon_0a^3\frac{\epsilon_r-1}{\epsilon_r+2}\mathbf E_0.}
$$
For example, write $\phi_{\rm in}=-B r\cos\theta$ and $\phi_{\rm out}=-E_0r\cos\theta+C\cos\theta/r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric <displacement> gives $\epsilon B=\epsilon_0(E_0+2C/a^3)$. Solving yields $B=3E_0/(\epsilon_r+2)$ and $C=a^3E_0(\epsilon_r-1)/(\epsilon_r+2)$. The $1/r^2$ potential is that of the <electric dipole moment> above. For $\epsilon_r=1$ all these induced effects vanish; as $\epsilon_r$ becomes very large the normal-field <sphere> and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.
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