Pre-Quantum Electrodynamics
The Field Inside a Dielectricemsm.esm.fpo.fid
Outside dielectric: OK, don't need microscopic details of field.
Inside dielectric: microscopic field is very complicated. Macroscopic field is what we should concentrate on. Defined as average field over region containing many constituents.
Suppose we want to calculate field at point \({\bf r}\) inside dielectric. Consider a sphere around \({\bf r}\), such that many atoms are within the sphere. Total macroscopic electric field: \[ {\bf E} = {\bf E}_{out} + {\bf E}_{in}. \] From Problem 3.47, average field (over a sphere) produced by charges outside sphere is equal to field they produce at center of sphere. Therefore, \({\bf E}_{out}\) is field at \({\bf r}\) from dipoles exterior to sphere, for which we can use \ref{Gr(4.9)}: \[ V_{out} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{out} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} {|{\bf r} - {\bf r}'|^3} \label{Gr(4.17)} \] For dipoles inside: cannot be treated in this fashion. But their average field is all we need, and this is given by \ref{Gr(3.105)} (see Example below) \[ {\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} \label{Gr(3.105)} \] irrespective of charge distribution within sphere.
\paragraph{Problem 3.47:} show that the average field inside sphere of radius \(R\) due to all charges within the sphere is \[ {\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{{\bf p}}{R^3} \label{Gr(3.105)} \] where \({\bf p}\) is the dipole moment. Many ways to do this. \paragraph{a)} show that average field due to single charge \(q\) at \({\bf r}\) inside sphere is same as field at \({\bf r}\) due to uniformly charged sphere with \(\rho = -q/(\frac{4}{3} \pi R^3)\), \[ \frac{1}{4\pi \varepsilon_0} \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' q \frac{{\bf r}' - {\bf r}}{|{\bf r} - {\bf r}'|^3}. \] \paragraph{Solution}: average field due to point charge \(q\) at \({\bf r}\): \[ {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' {\bf E} ({\bf r}'), \hspace{1cm} {\bf E}({\bf r}') = \frac{q}{4\pi \varepsilon_0} \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} \] so \[ {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \frac{q}{4\pi\varepsilon_0} \int d\tau' \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} \] Field at \({\bf r}\) due to uniformly charged sphere: \[ {\bf E}_s ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int d\tau' \rho({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \] If \(\rho = -q/(\frac{4}{3} \pi R^3)\), the two expressions coincide.
\paragraph{b)} Express the latter in terms of the dipole moment. \paragraph{Solution:} from Problem 2.12, field inside uniformly charged sphere is \[ {\bf E} ({\bf r}) = \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \] so \[ {\bf E}_{s} ({\bf r}) = \left. \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \right|_{\rho = \frac{-q}{\frac{4}{3}\pi R^3}} = -\frac{q}{4\pi\varepsilon_0} \frac{\hat{\bf r}}{R^3} = -\frac{{\bf p}}{4\pi\varepsilon_0 R^3}. \] \paragraph{c)} Arbitrary charge distribution \paragraph{Solution:} just use superposition ! \paragraph{d)} Show that the average field over the volume of a sphere due to all the charges outside is the same as the field they produce at the center: \paragraph{Solution:} Field at \({\bf r}'\) due to charge at \({\bf r}\) outside the sphere: \[ {\bf E} ({\bf r}') = \frac{q}{4\pi\varepsilon_0} \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} \] Averaged over sphere: \[ {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \frac{q}{4\pi\varepsilon_0} \int d\tau' \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} \] This is same as field produced at \({\bf r}\) outside sphere, by uniformly charged sphere with \(\rho = -q/(\frac{4}{3} \pi R^3)\). We know that this is simply \[ {\bf E}_s = \frac{1}{4\pi\varepsilon_0} (-q) \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \] where the sphere is centered on \({\bf r}'\). But this is precisely the field produced at the center of the sphere \({\bf r}'\), by a charge \(q\) sitting at \({\bf r}\).
We thus have \[ {\bf E}_{in} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} \] Since \({\bf p} = \frac{4}{3} \pi R^3 {\bf P}\), we get \[ {\bf E}_{in} = -\frac{1}{3\varepsilon_0} {\bf P} \label{Gr(4.18)} \]
Now: by assumption, \({\bf P}\) does not vary significantly over volume of sphere. Term left out of \ref{Gr(4.17)} is thus field at center of uniformly polarized sphere. But this is precisely equal to the \({\bf E}_{in}\) contribution, so macroscopic field is given by the potential \[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} {|{\bf r} - {\bf r}'|^3} \label{Gr(4.19)} \] where \({\cal V}\) is the entire volume of the dielectric.
Created: 2022-02-08 Tue 06:55