Work through the three derivations below in order; each uses the result of the one…

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Electric fields in matter

The Boundary-Value Problem, Ch Energy and force in dielectrics

Problem

Work through the three derivations below in order; each uses the result of the one before it. ((a)) A region contains linear dielectric media, D=εE\mathbf{D}=\varepsilon\mathbf{E}, and a set of conductors carrying free charge, with all fields vanishing at infinity. Free charge is brought in from infinity in small increments δρf\delta\rho_f, the surrounding dielectric settling into equilibrium at each stage. Starting from the electrostatic work done moving δρf\delta\rho_f against the potential VV already present, and using ∇⋅δD=δρf\nabla\cdot\delta\mathbf{D}=\delta\rho_f, derive δWfree=∫E⋅δD dτ\delta W_{\rm free}=\int\mathbf{E}\cdot\delta\mathbf{D}\,d\tau (the integral over all space). Then, holding the dielectric fixed (ε\varepsilon constant) so that only the field changes, integrate to Wfree=12∫D⋅E dτW_{\rm free}=\tfrac12\int\mathbf{D}\cdot\mathbf{E}\,d\tau. ((b)) A model molecule is a charge −qe-q_e bound to a fixed positive charge by an isotropic spring of force constant ksk_s. In a uniform applied field EE directed along the spring, the charge is pulled to a new equilibrium at displacement x0x_0, giving an induced dipole moment p=αEp=\alpha E with α=qe2/ks\alpha=q_e^2/k_s. Find the work WspringW_{\rm spring} stored in stretching the spring to x0x_0, and the interaction energy Uint=−pEU_{\rm int}=-pE of the resulting dipole with the applied field. Add the two and show that the molecule's total energy in the field is −12αE2-\tfrac12\alpha E^2. ((c)) A set of conductors is held at fixed potentials ViV_i by ideal batteries, with linear dielectrics filling the space between them, and the stored energy is W=12∑iQiViW=\tfrac12\sum_iQ_iV_i. One conductor is displaced quasi-statically through a small distance δx\delta x, its charge and the charges of the others changing so as to keep every ViV_i fixed. By computing the work the batteries must supply and balancing it against the change in stored energy and the mechanical work Fx δxF_x\,\delta x delivered by the field, show that Fx=+(∂W∂x)VF_x=+\left(\dfrac{\partial W}{\partial x}\right)_V.

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