##### The Field Inside a Dielectric
emsm.esm.po.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: \[ {\bf E} = {\bf E}_{out} + {\bf E}_{in}. \]

In p_ball_avg, we established that the average field (over a sphere) produced by charges outside the sphere is equal to the field they produce at center of sphere. Therefore, \({\bf E}_{out}\) is the field at \({\bf r}\) from dipoles exterior to the sphere, for which we can use p_P:

- Gr (4.17)

\[ \phi_{out} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{out} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} {|{\bf r} - {\bf r}'|^3} \tag{p_out}\label{p_out} \] The dipoles inside the sphere however cannot be treated in this fashion. However, their average field is all we need, and this is given by (see Example below for the derivation)

- Gr (3.105)

\[ {\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} \tag{Eavg_p}\label{Eavg_p} \] irrespective of charge distribution within sphere.

Average field inside a sphere

Task: show that the average field inside a sphere of radius \(R\) due to all charges within the sphere is \[ {\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{{\bf p}}{R^3} \] where \({\bf p}\) is the dipole moment, irrespective of what the charge distribution is.

Solution: there are many ways to do this.

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 \(\rho = -q/(\frac{4}{3} \pi R^3)\), \[ \frac{1}{4\pi \varepsilon_0} \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' q \frac{{\bf r}' - {\bf r}}{|{\bf r} - {\bf r}'|^3}. \]

Solution a): 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.

b) Express the latter in terms of the dipole moment.

Solution b): from E_uni_sph, the field inside a uniformly charged sphere is \[ {\bf E} ({\bf r}) = \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \] so \[ {\bf E}_{s} ({\bf r}) = \left. \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \right|_{\rho = \frac{-q}{\frac{4}{3}\pi R^3}} = -\frac{q}{4\pi\varepsilon_0} \frac{\hat{\bf r}}{R^3} = -\frac{{\bf p}}{4\pi\varepsilon_0 R^3}. \]

c) Generalize to an arbitrary charge distribution

Solution c): just use superposition!

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:

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 the same as the 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 \[ {\bf E}_{in} = -\frac{1}{3\varepsilon_0} {\bf P} \label{Gr(4.18)} \]

By assumption, \({\bf P}\) does not vary significantly over the volume of sphere. The term left out of p_out is thus the field at the center of a uniformly polarized sphere. But this is precisely equal to the \({\bf E}_{in}\) contribution, so macroscopic field is given by the potential

\[ \phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} {|{\bf r} - {\bf r}'|^3} \] where \({\cal V}\) is the entire volume of the dielectric.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42