Q146 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Physics
An electric dipole consists of two particle each of mass $1 \mathrm{~kg}$ separated by $\mathrm{l} \mathrm{m}$ carrying charges $1 \mu \mathrm{C}$ and $-1 \mu \mathrm{C}$ respectively. It is in equilibrium in a uniform electric field of $2 \times 10^4 \mathrm{Vm}^{-1}$. If it is deflected by a small angle $2^{\circ}$, minimum time taken by it to come back again to the mean position is (in seconds)
- Correct answerA$2.5 \pi$
- B$2 \pi$
- C$5 \pi$
- D$4 \pi$
Solution
Torque on dipole, when it is deflected by a small angle $\theta$ is $\tau=p E \sin \theta=p E \theta$ \{ for small angle, $\sin \theta=\theta\}$
But $\quad \tau=I \alpha$
where, $\quad I=m r^2+m r^2=2 m r^2$
So, angular acceleration of dipole is
$$
\alpha=\frac{p E \theta}{I}
$$
Time to align back with field is $\left(\right.$ using $\left.\theta=\frac{1}{2} \alpha t^2\right)$
$$
t=\sqrt{\frac{2 \theta}{\alpha}}=\sqrt{\frac{2 \theta I}{p E \theta}}=\sqrt{\frac{2 I}{P E}}=\sqrt{50} \approx 2 \cdot 5 \pi(\mathrm{s})
$$
Hence, the correct answer is option: 'A'.