A Dipole in an Electric Field — Question 10

Question 10

A torsion spring competes with an initially opposing electric field

A rigid dipole rotates about a fixed center. A torsion spring favors $\theta=0$ and has energy $\kappa\theta^2/2$, with $\kappa>0$ and angle in radians. The field is opposite the moment at $\theta=0$, giving total energy $U(\theta)=\kappa\theta^2/2+pE\cos\theta$.

1. Derive the equilibrium equation and classify the stability of $\theta=0$ for $pE<\kappa$, $pE>\kappa$, and equality.

2. Just above threshold, set $pE=\kappa(1+\epsilon)$ with $0<\epsilon\ll1$. Estimate the two small nonzero equilibria and their stability; evaluate the estimate at $\epsilon=.0200$.

Question 10 - Solution

A torsion spring competes with an initially opposing electric field

(a) Method and calculation

$$U\prime=\kappa\theta-pE\sin\theta=0,\quad U\prime\prime(0)=\kappa-pE.$$$$pE=\kappa:\quad U(\theta)-U(0)=\frac{\kappa\theta^4}{24}+O(\theta^6).$$The zero-angle equilibrium is stable below threshold, unstable above it, and stable with a leading quartic restoring energy at equality.

(b) Interpretation and further calculation

$$0\simeq(\kappa-pE)\theta+\frac{pE}{6}\theta^3\Rightarrow\theta_\pm\simeq\pm\sqrt{\frac{6\epsilon}{1+\epsilon}}.$$$$U\prime\prime(\theta_\pm)\simeq2(pE-\kappa)>0,\quad\theta_\pm\simeq\pm.3430\,\mathrm{rad}\;(\pm19.65^\circ).$$The spring-field competition creates two nearby stable orientations; these values are small-angle estimates, not exact roots.

Physical check

At exact threshold the linear oscillation frequency is zero, but the positive quartic energy still makes the zero-angle state a local minimum.