1. Calculating the Dipole Moment:

The ring has total charge $q$ and circumference $\approx 2\pi r$ (since $l \ll r$). The linear charge density is $\lambda = \frac{q}{2\pi r}$.

The “missing” charge at the gap is $q_{gap} = \lambda l = \frac{q l}{2\pi r}$.

The dipole moment $\vec{p}$ of the system is equivalent to a negative charge $-q_{gap}$ at the gap position and a positive charge $+q_{gap}$ at the center (balancing the rest of the ring). Thus, the magnitude of the dipole moment is:

$$ p = (q_{gap}) \times r = \left( \frac{q l}{2\pi r} \right) r = \frac{q l}{2\pi} $$

The vector $\vec{p}$ points from the gap towards the center of the ring.

2. Work-Energy Application:

Initially, the gap is “parallel to E”. This phrasing typically implies the tangent to the ring at the gap is parallel to E. If the gap is at the top (12 o’clock), the tangent is horizontal, and E is horizontal. In this position, the radius vector is vertical, so $\vec{p}$ is perpendicular to $\vec{E}$.

• Initial Potential Energy: $U_i = -\vec{p} \cdot \vec{E} = -pE \cos(90^\circ) = 0$.
• Final Potential Energy: The ring rotates until $\vec{p}$ aligns with $\vec{E}$ (stable equilibrium). Here $\theta = 0^\circ$. $U_f = -pE \cos(0^\circ) = -pE$.

According to the Conservation of Energy ($\Delta K + \Delta U = 0$):

$$ K_f – K_i = -(U_f – U_i) $$ $$ \frac{1}{2} I \omega^2 – 0 = -(-pE – 0) = pE $$

3. Solving for Angular Speed:

The moment of inertia of a thin ring is $I = m r^2$. Substituting values:

$$ \frac{1}{2} (m r^2) \omega^2 = \left( \frac{q l}{2\pi} \right) E $$ $$ m r^2 \omega^2 = \frac{q l E}{\pi} $$ $$ \omega^2 = \frac{q l E}{\pi m r^2} $$