Classical bound-charge model – polarizability of a harmonically bound charge A particle of charge Q (coulombs) is elastically bound to its equilibrium position with spring constant f (newtons per metre). Under a static electric field E, what is the polarizability α of this one-particle system?

Introduction / Context:
The classical Lorentz (harmonic oscillator) model treats an electron or ion bound by a restoring force F = −f x. In a static electric field E, the bound charge is displaced, forming an electric dipole. The proportionality between induced dipole moment and field is the polarizability α, a key material parameter in dielectrics and optics.

Given Data / Assumptions:

• Point charge Q bound by linear spring constant f.
• Quasi-static field (ignore inertia and damping).
• Dipole moment p = Q x for displacement x.

Concept / Approach:

Static force balance: electric force Q E is balanced by spring force f x. Solve for displacement x, then compute dipole moment p = Q x. The polarizability is defined by p = α E, so α is the factor relating dipole to field strength. This model underlies frequency-dependent polarizability as well, where α(ω) = Q^2 /(f − m ω^2 + j γ ω), but at ω = 0 it reduces to the simple static form.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional check: Q^2 / f has units C^2/(N/m) = C^2·m/N, consistent with dipole per unit field (C·m)/(V/m) = C·m^2/V.

Why Other Options Are Wrong:

Q/f and f/Q^2 have incorrect dimensions; Q f and Q^2 f are not polarizabilities and grow with stiffness rather than decreasing, which is unphysical.

Common Pitfalls:

Forgetting that the induced dipole scales with Q^2 (one Q from force, one from dipole definition) and inversely with stiffness f.

Final Answer:

α = Q^2 / f