Pre-Quantum Electrodynamics

Polarized Objects; Bound Charges emsm.esm.po

What is the electric field produced by an object with polarization \({\bf P}\)? Let us try to address this question by Working with the potential. For a single dipole, we have p_di \[ \phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{({\bf r} - {\bf r}') \cdot {\bf p}}{|{\bf r} - {\bf r}'|^3} \label{Gr(4.8)} \] Considering a dipole moment per unit volume \({\bf P}\) (our definition of the polarization), we get

  • Gr (4.9)

\[ \phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|^3} \tag{p_P}\label{p_P} \] This equation is perfectly acceptable and usable as it is. There exists however another convenient representation. Starting from \[ {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \] we can write (using product rule div_prod)

\begin{align*} \phi({\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

\begin{equation*} \phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} d{\bf a}' \cdot \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|} - \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|}. \end{equation*}

Interpretation: first terms is like contribution of a surface charge,

  • Gr (4.11)

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

and second term looks like contribution of a volume charge,

  • Gr (4.12)

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

Using these definitions,

  • Gr (4.13)

\[ \phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} da' \frac{\sigma_b ({\bf r}')}{|{\bf r} - {\bf r}'|} + \frac{1}{4\pi \varepsilon_0} \int_{\cal V} d\tau' \frac{\rho_b ({\bf r}')}{|{\bf r} - {\bf r}'|}. \tag{p_bound}\label{p_bound} \]

These bound charges faithfully represent the object's sources for electrical fields.

Example: uniformly polarized sphere

Task: compute the electric field produced by a uniformly polarized sphere of radius \(R\).

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): \[ \phi(r, \theta) = \left\{ \begin{array}{cc} \frac{P}{3\varepsilon_0} r\cos \theta, & r \leq R \\ \frac{P}{3\varepsilon_0} \frac{R^3}{r^2} \cos \theta, & r \geq R. \end{array} \right. \]

But \(r\cos \theta = z\), so the field inside the sphere is uniform,

\[ {\bf E} = -{\boldsymbol \nabla} \phi = -\frac{P}{3\varepsilon_0} \hat{\bf z} = -\frac{1}{3\varepsilon_0} {\bf P}, \hspace{1cm} r < R. \] Outside the sphere, the potential is identical to that of pure point dipole at origin,

\[ \phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \frac{{\bf p} \cdot {\hat {\bf r}}}{r^2}, \hspace{1cm} r > R \] where the total dipole moment is simply the integral over the polarization,

\[ {\bf p} = \frac{4}{3} \pi R^3 {\bf P} \]

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Author: Jean-Sébastien Caux

Created: 2024-02-27 Tue 10:31