Pre-Quantum Electrodynamics

• PM 10
• Gr 4.4.3

To charge a capacitor: \[ W = \frac{1}{2} C V^2 \] If capacitor filled with linear isotropic dielectric: \[ C = \varepsilon_r C_{vac} \] From electrostatics: \[ W = \frac{\varepsilon_0}{2} \int d\tau E^2 \] How is this changed? Counting only energy in fields: should decrease by factor \(1/\varepsilon_r^2\). However, this would neglect the strain energy associated to the distortion of the polarized constituents of the dielectric medium.

Derivation from scratch: bring in free charge: \(\rho_f\) increased by \(\Delta \rho_f\), polarization changes (also bound charge distribution). Work done on free charges (only that matters): \[ \Delta W = \int d\tau (\Delta \rho_f ({\bf r})) \phi ({\bf r}). \label{Gr(4.56)} \] But \(\rho_f = {\boldsymbol \nabla} \cdot {\bf D}\) so \(\Delta \rho_f = {\boldsymbol \nabla} \cdot (\Delta {\bf D})\), so \[ \Delta W = \int d\tau ({\boldsymbol \nabla} \cdot (\Delta {\bf D})) \phi = \int d\tau {\boldsymbol \nabla} \cdot ((\Delta {\bf D}) \phi) + \int d\tau (\Delta {\bf D}) \cdot {\bf E} \] First integral: divergence theorem changes it to a surface integral which vanishes when integrating over all space. Therefore, \[ \Delta W = \int d\tau (\Delta {\bf D}) \cdot {\bf E} \label{Gr(4.57)} \] This applies to any material.

Special case of linear isotropic dielectric: \({\bf D} = \varepsilon {\bf E}\), so \[ \Delta W = \Delta \left( \frac{1}{2} \int d\tau {\bf D} \cdot {\bf E} \right) \] Total work done: