Consider an electric dipole comprising two charges +q and -q each with mass m, separated by a fixed distance d and initially at rest with its dipole moment pointing along î. A uniform electric field E ĵ is turned on at time t = 0 and it is turned off at t = t_f, when the dipole moment makes an angle θ_f with î. Neglecting any sources of energy loss, correct option(s) is/are:

The correct options are:
1. If the magnitude of the final angular velocity ${\omega }_f=\sqrt{\frac{2qE}{md}}$, then ${\theta }_f=\frac{\pi }{6}$.
2. For θ_f = π/4, the dipole rotates around its center of mass with a constant angular velocity after t > t_f.

Step 1: Analyzing the Forces and Motion of the Center of Mass
The electric dipole consists of two equal and opposite charges, $+q$ and $-q$.
When a uniform electric field $\overrightarrow{E}=E\hat{j}$ is applied:
Force on positive charge, ${\overrightarrow{F}}_{+}=+qE\hat{j}$
Force on negative charge, ${\overrightarrow{F}}_{-}=-qE\hat{j}$
The net force acting on the dipole is:

${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{+}+{\overrightarrow{F}}_{-}=qE\hat{j}-qE\hat{j}=\overrightarrow{0}$

Since the net force acting on the center of mass is zero at all times, the center of mass remains at rest throughout the motion and is not deflected.

Step 2: Moment of Inertia of the Dipole
Both charges of mass $m$ are located at a distance $\frac{d}{2}$ from the center of mass.
The moment of inertia $I$ about the axis passing through the center of mass and perpendicular to the plane of rotation is:

$I=m{\left(\frac{d}{2}\right)}^2+m{\left(\frac{d}{2}\right)}^2=2m\frac{d^2}{4}=\frac{1}{2}m{d}^2$

Step 3: Work-Energy Theorem for Rotation
The electric dipole moment vector is $\overrightarrow{p}=qd\hat{i}$ initially, making an angle $\theta =0$ with the x-axis ($\hat{i}$).
The electric field vector is directed along the y-axis ($\hat{j}$), so the angle between $\overrightarrow{p}$ and $\overrightarrow{E}$ is initially $\frac{\pi }{2}$.
When the dipole rotates by an angle ${\theta }_f$ with respect to $\hat{i}$, the angle between $\overrightarrow{p}$ and $\overrightarrow{E}$ becomes $\left(\frac{\pi }{2}-{\theta }_f\right)$.
The potential energy of a dipole in a uniform electric field is given by:

$U=-\overrightarrow{p}·\overrightarrow{E}=-pE\cos \varphi$

Initial potential energy ($\theta =0$, $\varphi =\frac{\pi }{2}$):

$U_i=-pE\cos \left(\frac{\pi }{2}\right)=0$

Final potential energy at ${\theta }_f$:

$U_f=-pE\cos \left(\frac{\pi }{2}-{\theta }_f\right)=-pE\sin {\theta }_f=-qEd\sin {\theta }_f$

By conservation of mechanical energy:

$K_i+U_i=K_f+U_f$

Since the dipole starts from rest, $K_i=0$. Thus:

$0+0=\frac{1}{2}I{\omega }_f^2-qEd\sin {\theta }_f$

$\frac{1}{2}\left(\frac{1}{2}m{d}^2\right){\omega }_f^2=qEd\sin {\theta }_f$

$\frac{1}{4}m{d}^2{\omega }_f^2=qEd\sin {\theta }_f\Rightarrow {\omega }_f^2=\frac{4qE\sin {\theta }_f}{md}$

Step 4: Verification of Options
For option 2: Given ${\omega }_f=\sqrt{\frac{2qE}{md}}$

${\left(\sqrt{\frac{2qE}{md}}\right)}^2=\frac{4qE\sin {\theta }_f}{md}$

$\frac{2qE}{md}=\frac{4qE\sin {\theta }_f}{md}\Rightarrow \sin {\theta }_f=\frac{1}{2}\Rightarrow {\theta }_f=\frac{\pi }{6}$

Therefore, option 2 is correct.

For option 4: At time $t=t_f$, the electric field is turned off ($E=0$).
When $E=0$, no torque acts on the dipole ($\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}=\overrightarrow{0}$).
According to angular momentum conservation, the angular acceleration becomes zero, and the dipole continues to rotate around its center of mass with a constant angular velocity for $t>t_f$.
Therefore, option 4 is also correct.