JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 13)
A dipole comprises of two charged particles of identical magnitude $$q$$ and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance '$$l$$'. If the dipole is placed in a uniform electric field '$$\bar{E}$$'; in such a way that dipole axis makes a very small angle with the electric field, '$$\bar{E}$$'. The angular frequency of the oscillations of the dipole when released is given by:
$$\sqrt{\frac{3 q E}{2 m l}}$$
$$\sqrt{\frac{4 q E}{m l}}$$
$$\sqrt{\frac{8 q E}{3 m l}}$$
$$\sqrt{\frac{8 q E}{m l}}$$
Explanation
_6th_April_Evening_Shift_en_13_1.png)
If released, it will oscillate about centre of mass.
For small ' $\theta$ '
$$ \begin{aligned} & \tau=-\mathrm{PE} . \theta \\\\ & \Rightarrow\left[2 \mathrm{~m} \frac{1^2}{9}+\mathrm{m} \frac{4 \mathrm{l}^2}{9}\right] \alpha=-\mathrm{qlE} \cdot \theta \\\\ & \Rightarrow \frac{2 \mathrm{ml}^2}{3} \alpha=-\mathrm{qlE} \cdot \theta \Rightarrow \alpha=-\frac{3 \mathrm{qE}}{2 \mathrm{ml}} \theta \\\\ & \omega=\sqrt{\frac{3 \mathrm{qE}}{2 \mathrm{ml}}} \end{aligned} $$
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