Question

An electric dipole consists of two particle each of mass $1kg$ separated by $1m$ carrying charges $1\mu C$ and $-1\mu C$ respectively. It is in equilibrium in a uniform electric field of $2\times 10^{4}V{m}^{-1}$. 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 $\tau =pE\sin \theta =pE\theta \{$ for small angle, $\sin \theta =\theta \}$
But $\tau =I\alpha$
where, $I=m{r}^2+m{r}^2=2m{r}^2$
So, angular acceleration of dipole is
$\alpha =\frac{pE\theta }{I}$
Time to align back with field is $($ using $\theta =\frac{1}{2}\alpha t^2)$
$t=\sqrt{\frac{2\theta }{\alpha }}=\sqrt{\frac{2\theta I}{pE\theta }}=\sqrt{\frac{2I}{PE}}=\sqrt{50}\approx 2\cdot 5\pi (s)$