An electric dipole of mass m , charge q , and length l is placed in a uniform electric field \vecE = E_0\hati . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

Electrostatics PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q59 with Solution

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Correct answer: B

A dipole of charges $+q$ and $-q$ separated by distance $l$ has dipole moment $p = ql.$

When it makes a small angle $\theta$ with the electric field, the restoring torque is $\tau = -pE_0\sin\theta = -q l E_0 \sin\theta.$

For small oscillations, $\sin\theta \approx \theta$ , so $\tau \approx -q l E_0\,\theta.$

This is the form of restoring torque for angular SHM.

For rotational motion, $I\,\ddot{\theta} = \tau.$

So, $I\,\ddot{\theta} = -q l E_0\,\theta,$ which gives $\ddot{\theta} + \frac{q l E_0}{I}\theta = 0.$

Hence the angular frequency is $\omega = \sqrt{\frac{q l E_0}{I}}.$

Therefore, $T = 2\pi \sqrt{\frac{I}{q l E_0}}.$

The dipole consists of two particles, each of mass $m/2$ (since total mass is $m$ ), separated by distance $l$ .

About its center, each mass is at distance $l/2$ .

Thus,

= m\cdot \frac{l^2}{4}.$$ So, $$I = \frac{m l^2}{4}.$$ --- 4. **Substitute into time period formula** $$T = 2\pi \sqrt{\frac{I}{q l E_0}} = 2\pi \sqrt{\frac{m l^2/4}{q l E_0}} = 2\pi \sqrt{\frac{m l}{4 q E_0}}.$$ This simplifies to $$T = \pi \sqrt{\frac{m l}{q E_0}}.$$ --- 5. **Check options** Given options are: - A: $\frac{1}{2\pi} \sqrt{\frac{ml}{2qE_0}}$ - B: $2\pi \sqrt{\frac{ml}{2qE_0}}$ - C: $2\pi \sqrt{\frac{ml}{qE_0}}$ - D: $\frac{1}{2 \pi} \sqrt{\frac{2 m l}{q E_0}}$ Our derived result is $$T = \pi \sqrt{\frac{m l}{q E_0}},$$ which does **not** match any option exactly. --- 6. **Possible convention used in the options** Many textbook problems assume the dipole consists of two point masses each of mass $m$ (so total mass $=2m$), while the statement says dipole of mass $m$ ambiguously. If each charge has mass $m$, then $$I = 2m\left(\frac{l}{2}\right)^2 = \frac{m l^2}{2}.$$ Then $$T = 2\pi \sqrt{\frac{I}{q l E_0}} = 2\pi \sqrt{\frac{m l^2/2}{q l E_0}} = 2\pi \sqrt{\frac{m l}{2 q E_0}},$$ which matches **Option B**. So the stored answer corresponds to interpreting $m$ as the mass of each particle, not total mass. --- 7. **Conclusion** Using the standard exam interpretation intended by the options, the correct answer is: $$\boxed{B}$$