A light insulating rod AB, 20 cm long, has a small ball fixed at each end…

Step 1. Dipole moment (a). $p = q\,l = 2.0 \times 10^{-6}\ \text{C} \times 0.20\ \text{m} = 4.0 \times 10^{-7}\ \text{C m}$, pointing from $B$ to $A$.

Step 2. Torque (b). $\tau = pE\sin60^\circ = 4.0 \times 10^{-7}\ \text{C m} \times 5.0 \times 10^{4}\ \text{N/C} \times 0.866 = 0.0173\ \text{N m}$. Taking moments of the two forces about $O$ gives the same, because the forces are a couple.

Step 3. Work to turn it round (c). The energy is $U = -pE\cos\theta$. From $\theta = 0$ to $180^\circ$ it rises by $2pE = 0.040\ \text{J}$.

Step 4. Small oscillations (d). About $O$, $I = m(0.05\ \text{m})^2 + m(0.15\ \text{m})^2 = 1.25 \times 10^{-4}\ \text{kg m}^2$. For a small angle, $\tau \approx -pE\,\theta$, so

Step 5. Angular speed (e). From $90^\circ$ to $0$ the energy falls by $pE$: $\tfrac12 I\omega^2 = pE$, so $\omega = \sqrt{2pE/I} = 17.89\ \text{rad/s}$.

Step 6. The axle's force (f). As the rod lines up, the torque is zero, so it has no angular acceleration. The centre of mass, the middle of the rod, is $0.05\ \text{m}$ from $O$ and moves in a circle. The electric forces add to zero, so the axle alone gives the centripetal force: