# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

#### The Electric Displacement
emsm.esm.D

Field due to polarization: effectively comes from bound surface and volume charges,

\[ \rho_b = -{\boldsymbol \nabla} \cdot {\bf P}, \hspace{1cm} \sigma_b = {\bf P} \cdot \hat{\bf n}. \]

Rest of charges: {\bf free charge} (electrons in conductors, ions embedded in dielectric, {\it i.e.} any charge which doesn't come from polarization).

Within dielectric: \[ \rho = \rho_b + \rho_f \label{Gr(4.20)} \] Gauss's law: \[ \varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E} = \rho = \rho_b + \rho_f = -{\boldsymbol \nabla} \cdot {\bf P} + \rho_f. \] Convenient way of writing: \[ {\boldsymbol \nabla} \cdot (\varepsilon_0 {\bf E} + {\bf P}) = \rho_f. \] Defining the

Electric displacement \({\bf D}\) \[ {\bf D} \equiv \varepsilon_0 {\bf E} + {\bf P} \label{Gr(4.21)} \]

Gauss's law becomes

\[ {\boldsymbol \nabla} \cdot {\bf D} = \rho_f \label{Gr(4.22)} \]

or in integral form

\[ \oint d{\bf a} \cdot {\bf D} = Q_{f_{enc}} \label{Gr(4.23)} \]

\paragraph{Example 4.4:} long straight wire, uniform line charge \(\lambda\), surrounded by rubber insulation to radius \(a\). Find \({\bf D}\). \paragraph{Solution:} cylindrical Gaussian surface, radius \(s\) and length \(L\). Applying \ref{Gr(4.23)}, \[ D 2\pi s L = \lambda L, \] so \[ {\bf D} = \frac{\lambda}{2\pi s} \hat{\bf s} \label{Gr(4.24)} \] This holds within insulation and outside of it. Outside, \({\bf P} = 0\) so \[ {\bf E} = \frac{1}{\varepsilon_0} {\bf D} = \frac{\lambda}{2\pi \varepsilon_0 s} \hat{\bf s}, \hspace{1cm} s > a. \] Inside: can't know the electric field, since \({\bf P}\) isn't known.

\paragraph{Note of caution:} while \ref{Gr(4.23)} might lead you to think that \[ {\bf D} ({\bf r}) = \frac{1}{4\pi} \int_{\cal V} d\tau' \rho_f ({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \] but this is {\bf not} correct. In particular, the curl of the displacement isn't always zero, \[ {\boldsymbol \nabla} \times {\bf D} = \varepsilon_0 ({\boldsymbol \nabla} \times {\bf E} + {\boldsymbol \nabla} \times {\bf P}) = {\boldsymbol \nabla} \times {\bf P} \label{Gr(4.25)} \] There is not 'potential' for \({\bf D}\).

##### In this section:

- Boundary Conditions emsm.esm.D.bc

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Author: Jean-Sébastien Caux

Created: 2022-02-09 Wed 22:40