Question

Two identical masses, of mass $m$ each and carrying equal and opposite charges of magnitude $q$, are attached to the ends of a non-conducting uniform rod, of mass $m$ and length $l$. The system is related from rest in uniform electric field, from $\theta ={53}^{\circ }$ as shown in figure. The angular acceleration of the rod, just after release is,

• A. $\frac{48qE}{35ml}$
• B. $\frac{28qE}{25ml}$
• C. $\frac{18qE}{7ml}$
• D. $\frac{qE}{5ml}$

Answer 1

The correct option is A $\frac{48qE}{35ml}$

The moment of inertia of the entire system about the axis of rotation, i.e. centre of dipole is,

$I=I_1+I_2+I_3$

$\Rightarrow I=m{\left(\frac{l}{2}\right)}^2+m{\left(\frac{l}{2}\right)}^2+\left(\frac{m{l}^2}{12}\right)$

$=\frac{7m{l}^2}{12}$

The torque on the dipole in the position shown in the diagram,
i.e. $\theta ={53}^{\circ }$, is,

${\tau }_{net}=pEsin\theta =pE\sin {53}^{\circ }=\frac{4}{5}pE$

$\Rightarrow I\alpha =\frac{4}{5}pE\Rightarrow \frac{7m{l}^2}{12}\alpha =\frac{4}{5}qlE$

$\therefore \alpha =\frac{48qE}{35ml}$

Hence, option $\left(A\right)$ is the correct answer.