#### 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: free charge (electrons in conductors, ions embedded in dielectric, 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 \({\bf D}\)

- Gr (4.21)

\[ {\bf D} \equiv \varepsilon_0 {\bf E} + {\bf P} \tag{D}\label{D} \]

Gauss's law becomes

- Gr (4.22)

\[ {\boldsymbol \nabla} \cdot {\bf D} = \rho_f \tag{GlD}\label{GlD} \]

or in integral form

- Gr (4.23)

\[ \oint d{\bf a} \cdot {\bf D} = Q_{f_{enc}} \tag{GlD_int}\label{GlD_int} \]

Example: long insulated wire

Consider a long straight wire, with uniform line charge \(\lambda\), surrounded by rubber insulation to radius \(a\).

Task: find \({\bf D}\).

Solution: cylindrical Gaussian surface, radius \(s\) and length \(L\). Applying GlD_int, \[ D 2\pi r L = \lambda L, \] so

\[ {\bf D} = \frac{\lambda}{2\pi r} \hat{\bf r} \] This holds within insulation and outside of it. Outside, \({\bf P} = 0\) so \[ {\bf E} = \frac{1}{\varepsilon_0} {\bf D} = \frac{\lambda}{2\pi \varepsilon_0 r} \hat{\bf r}, \hspace{1cm} r > a. \] Inside: can't know the electric field, since \({\bf P}\) isn't known.

Warning: while GlD_int 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 not correct, because the curl of the displacement isn't necessarily zero,

- Gr (4.25)

\[ {\boldsymbol \nabla} \times {\bf D} = \varepsilon_0 ({\boldsymbol \nabla} \times {\bf E} + {\boldsymbol \nabla} \times {\bf P}) = {\boldsymbol \nabla} \times {\bf P} \tag{curlDcurlP}\label{curlDcurlP} \] There is therefore no 'potential' for \({\bf D}\).

##### In this section:

- Boundary Conditions emsm.esm.D.bc

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42