Question

An electric dipole consists of two particle each of mass $1\, kg$ separated by $1\, m$ carrying charges $1 \mu C$ and $- 1 \mu C$ respectively. It is in equilibrium in a uniform electric field of $2 \times 10^{4} Vm ^{-1}$. 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)

• A. $2.5\, \pi $
• B. $2\, \pi $
• C. $5\, \pi $
• D. $4\, \pi $

Answer 1

Answer: (A)

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 $\tau=I \alpha$
where, $I=m r^{2}+m r^{2}=2 m r^{2}$
So, angular acceleration of dipole is
$\alpha=\frac{p E \theta}{I}$
Time to align back with field is $\left(\right.$ using $\left.\theta=\frac{1}{2} \alpha t^{2}\right)$
$t=\sqrt{\frac{2 \theta}{\alpha}}=\sqrt{\frac{2 \theta I}{p E \theta}}=\sqrt{\frac{2 I}{P E}}=\sqrt{50} \approx 2 \cdot 5 \pi( s )$