= Solution
Direct variation of the <Lagrangian> gives
$$
m\ddot\rho-m\rho\dot\theta^2+k\rho-qE\cos\theta-qB\rho\dot\theta=0,
$$
$$
m\rho^2\ddot\theta+2m\rho\dot\rho\dot\theta
+qB\rho\dot\rho+qE\rho\sin\theta=0.
$$
For a static configuration, $k\rho=qE\cos\theta$ and $qE\rho\sin\theta=0$. Since $qE>0$ and the radius is positive,
$$
\boxed{\rho_0=qE/k,\qquad\theta_0=0\pmod{2\pi}.}
$$
The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_\rho=0$ but $p_\theta=qB\rho_0^2/2$, rather than zero canonical <angular momentum>.
Back to article page
