Pre-Quantum Electrodynamics

The Field of a Polarized Object emsm.esm.fpo

For a macroscopic moment: local induced/permanent dipole moments tend to somewhat cancel each other, leaving a residual polarization. Convenient measure: \[ {\bf P} \equiv ~\mbox{dipole moment per unit volume} \] called the {\bf polarization}. What is the electric field produced by an object with polarization \({\bf P}\) ? Work with the potential. For a single dipole, \ref{eq:electric_dipole}: \[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{({\bf r} - {\bf r}') \cdot {\bf p}}{|{\bf r} - {\bf r}'|^3} \label{Gr(4.8)} \] With dipole moment per unit volume \({\bf P}\), we get \[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(4.9)} \] This is correct as it is. There exists however another convenient representation. We know that \[ {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \] so we can write (using product rule number 5, ${\boldsymbol ∇} ⋅ (f {\bf A}) = f ({\boldsymbol ∇} ⋅ {\bf A}) + {\bf A} ⋅ {\boldsymbol ∇} f$~)

\begin{align} V({\bf r}) &= \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' {\bf P} ({\bf r}') \cdot {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} \nonumber \\ & = \frac{1}{4\pi \varepsilon_0} \left[ \int_{\cal V} d\tau' {\boldsymbol \nabla}' \cdot \left( \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right) - \int_{\cal V} d\tau' \frac{1}{|{\bf r} - {\bf r}'|} {\boldsymbol \nabla}' \cdot {\bf P} ({\bf r}') \right] \end{align}

Using the divergence theorem, this becomes \[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} d{\bf a}' ⋅ \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|}

\label{Gr(4.10)} \] Interpretation: first terms is like contribution of a surface charge,

\[ \sigma_b({\bf r}) = {\bf P} ({\bf r}) \cdot \hat{\bf n} \label{Gr(4.11)} \]

and second term looks like contribution of a volume charge,

\[ \rho_b ({\bf r}) = -{\boldsymbol \nabla} \cdot {\bf P} ({\bf r}) \label{Gr(4.12)} \]

Using these definitions,

\[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} d{\bf a}' ⋅ \frac{σ_b ({\bf r}')}{|{\bf r} - {\bf r}'|}

  • \frac{1}{4\pi \varepsilon_0} ∫_{\cal V} dτ' \frac{ρ_b ({\bf r}')}{|{\bf r} - {\bf r}'|}.

\label{Gr(4.13)} \]

These {\bf bound charges} faithfully represent the object's sources for electrical fields.

\paragraph{Example 4.2:} electric field produced by uniformly polarized sphere of radius \(R\). \paragraph{Solution:} put \(z\) axis along \({\bf P}\). Since \({\bf P}\) is uniform, \(\rho_b = 0\). Surface charge: \[ \sigma_b ({\bf r}) = {\bf P} \cdot \hat{\bf n} = P \cos \theta. \] This was computed in Example: surface charge density on a sphere (eq. p_uni_ch_sph): \[ V(r, \theta) = \left\{ \begin{array}{cc} \frac{P}{3\varepsilon_0} r\cos \theta, & r \leq R \\ \frac{P}{3\varepsilon_0} \frac{R^3}{r^2} \cos \theta, & r \geq R. \end{array} \right. \] But \(r\cos \theta = z\), so the field inside the sphere is uniform, \[ {\bf E} = -{\boldsymbol \nabla} V = -\frac{P}{3\varepsilon_0} \hat{\bf z} = -\frac{1}{3\varepsilon_0} {\bf P}, \hspace{1cm} r < R. \label{Gr(4.14)} \] Outside sphere, potential is identical to that of pure point dipole at origin, \[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \frac{{\bf p} \cdot {\hat {\bf r}}}{r^2}, \hspace{1cm} r > R \label{Gr(4.15)} \] where the total dipole moment is simply the integral over the polarization, \[ {\bf p} = \frac{4}{3} \pi R^3 {\bf P} \label{Gr(4.16)} \]

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Author: Jean-Sébastien Caux

Created: 2022-02-14 Mon 20:35