A Dipole in an Electric Field — Question 4

Section 22-8 · Physics pages 3618–3619

Question 4 Small angular oscillations and anti-aligned instability share one time scale A freely rotating dipole with moment p and central moment of inertia I is placed in a constant uniform field E. Its angular equation is I ¨θ = −pE sin θ. No damping is present. E aligned anti-aligned Schematic; labels, not drawing lengths, specify the data. (a) Linearize near alignment, derive the small-oscillation period and state how doubling E changes it. (b) Linearize near anti-alignment by writing θ = π + ϕ. Solve for a small displacement ϕ(0) = ϕ0, ˙ϕ(0) = 0 while the linear approximation holds.
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Question 4 - Solution Small angular oscillations and anti-aligned instability share one time scale E aligned anti-aligned (a) Method and calculation |θ| ≪1 : ¨θ + Ω2θ = 0, Ω= √ pE/I. T = 2π √ I/(pE), T(2E) = T(E)/ √ 2. (b) Interpretation and further calculation sin(π + ϕ) ≃−ϕ ⇒¨ϕ −Ω2ϕ = 0. ϕ(t) = ϕ0 cosh Ωt. Alignment is stable under small angular perturbations; anti-alignment has zero torque exactly but amplifies any small displacement. Physical check The linear unstable solution ceases to be accurate once |ϕ| is no longer small.

Question 4

Small angular oscillations and anti-aligned instability share one time scale

A freely rotating dipole with moment p and central moment of inertia I is placed in a constant uniform field E. Its angular equation is Iθ̈=−pEsin⁡θI\ddot\theta=-pE\sin\theta. No damping is present.

Diagram shown in the original SVG worksheet.

  1. Linearize near alignment, derive the small-oscillation period and state how doubling E changes it.

  2. Linearize near anti-alignment by writing θ=π+ϕ\theta=\pi+\phi. Solve for a small displacement ϕ(0)=ϕ0\phi(0)=\phi_0, ϕ̇(0)=0\dot\phi(0)=0 while the linear approximation holds.

Question 4 - Solution

Small angular oscillations and anti-aligned instability share one time scale

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|θ|≪1:θ̈+Ω2θ=0,Ω=pE/I.|\theta|\ll1:\quad\ddot\theta+\Omega^2\theta=0,\quad\Omega=\sqrt{pE/I}.T=2πI/(pE),T(2E)=T(E)/2.T=2\pi\sqrt{I/(pE)},\quad T(2E)=T(E)/\sqrt2.

(b) Interpretation and further calculation

sin⁡(π+ϕ)≃−ϕ⇒ϕ̈−Ω2ϕ=0.\sin(\pi+\phi)\simeq-\phi\Rightarrow\ddot\phi-\Omega^2\phi=0.ϕ(t)=ϕ0cosh⁡Ωt.\phi(t)=\phi_0\cosh\Omega t.Alignment is stable under small angular perturbations; anti-alignment has zero torque exactly but amplifies any small displacement.

Physical check

The linear unstable solution ceases to be accurate once |ϕ||\phi| is no longer small.