Pre-Quantum Electrodynamics
Susceptibility, Permittivity, Dielectric Constantemsm.esm.ld.sp
- PM 10
- Gr 4.4.1
For many substances: polarization is proportional to field, at least if the latter isn't too strong:
Constant \(\chi_e\): called the electric susceptibility of the medium. Since \(\varepsilon_0\) is there, \(\chi_e\) is dimensionless. Materials that obey PchiE are called linear dielectrics.
Note: \({\bf E}\) on the RHS of PchiE is the total electric field, due to free charges and to the polarization itself. Putting a dielectric in field \({\bf E}_0\), we can't compute \({\bf P}\) directly from PchiE.
In linear dielectrics: \[ {\bf D} = \varepsilon_0 {\bf E} + {\bf P} = \varepsilon_0 {\bf E} + \varepsilon_0 \chi_e {\bf E} = \varepsilon_0 (1 + \chi_e) {\bf E} \label{Gr(4.31)} \] so
where \(\varepsilon\) is called the permittivity of the material. In vacuum, susceptibility is zero, permittivity is \(\varepsilon_0\). Also,
\[ \varepsilon_r \equiv 1 + \chi_e = \frac{\varepsilon}{\varepsilon_0} \tag{epsr}\label{epsr} \] is called the relative permittivity or dielectric constant of the material. This is all just nomenclature, everything is already in PchiE.
Example: metal sphere with insulator shell
Consider a metal sphere of radius \(a\) carrying charge \(Q\), surrounded out to radius \(b\) by a linear dielectric material of permittivity \(\varepsilon\).
Task: find the potential at the center (relative to infinity).
Solution: we need to know \({\bf E}\). Could try to locate bound charge: but we don't know \({\bf P}\)! What we do know: free charge, situation is spherically symmetric, so can calculate \({\bf D}\) using GlD_int: \[ {\bf D} = \frac{Q}{4\pi r^2} \hat{\bf r}, \hspace{1cm} r > a. \] Inside sphere, \({\bf E} = {\bf P} = {\bf D} = 0\). Find \({\bf E}\) using DepsE: \[ {\bf E} = \left\{ \begin{array}{cc} \frac{Q}{4\pi \varepsilon r^2} \hat{\bf r}, & a < r < b, \\ \frac{Q}{4\pi \varepsilon_0 r^2} \hat{\bf r}, & r > b. \end{array} \right. \] The potential is thus \[ \phi = -\int_\infty^0 d{\bf l} \cdot {\bf E} = -\int_\infty^b dr \frac{Q}{4\pi\varepsilon_0 r^2} - \int_b^a dr \frac{Q}{4\pi\varepsilon r^2} = \frac{Q}{4\pi} \left( \frac{1}{\varepsilon_0 b} + \frac{1}{\varepsilon a} - \frac{1}{\varepsilon b} \right). \]
It was thus not necessary to compute the polarization or the bound charge explicitly. We can however extract those from the polarization, \[ {\bf P} = \varepsilon_0 \chi_e {\bf E} = \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon r^2} \hat{\bf r}, \] giving \[ \rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = 0, \hspace{1cm} \sigma_b = {\bf P} \cdot \hat{\bf n} = \left\{ \begin{array}{cc} \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon b^2}, & \mbox{outer surface} \\ -\frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon a^2}, & \mbox{inner surface} \end{array} \right. \]
A dielectric is thus like an imperfect conductor: the charge \(Q\) is not fully screened.
In linear dielectrics, the parallel between \({\bf E}\) and \({\bf D}\) is also not perfect.
Remark: since \({\bf P}\) and \({\bf D}\) are both proportional to \({\bf E}\) inside the dielectric, does it mean that their curl vanishes like for \({\bf E}\)? No: if there is a boundary between two materials with different dielectric constants, then a closed loop integral of e.g. \({\bf P}\) would not vanish.
The only situation in which this parallel works is when space is entirely filled with a homogeneous linear dielectric, but then this is just like a vacuum with a different permittivity.
Example: parallel-plate capacitor filled with dielectric material
Consider a parallel-plate capacitor filled with insulating material of dielectric constant \(\varepsilon_r\).
Task: determine the effect on the capacitance.
Solution: field confined between plates, and reduced by factor \(1/\varepsilon_r\). Potential difference \(V\) also reduced by same factor. Since \(Q = C/V\), capacitance is increased by factor of \(\varepsilon_r\), so
\[ C = \varepsilon_r C_{vac} \]
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Created: 2024-02-27 Tue 10:31