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As shown in the figure, a particle A of mass $2\,m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q.$ The system is placed in an electric field $\vec E$ . The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E $ . After the system settles into equilibrium, one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find maximum tension in the rod.
$qE+qE\theta_0^2$
$qE+\frac{qE\theta_0^2}{4}$
$qE+\frac{qE\theta_0^2}{3}$
$qE+\frac{qE\theta_0^2}{6}$
Solution
$I=\frac{m u \ell^{2}}{9}+\frac{2 m \ell^{2}}{9}$
$I=\frac{6 m \ell^{2}}{9}$
${\tau _{{\rm{net}}}} = qE\ell \sin \theta $
${\tau _{{\rm{net}}}} = I\alpha $
for small oscillation $\alpha=\omega^{2} \theta$
${\rm{qE}}\ell \theta = \frac{{2{\rm{m}}{\ell ^2}}}{3}\alpha \Rightarrow \frac{{3{\rm{qE}}\theta }}{{2{\rm{m}}\ell }} = \omega _{phase}^2$
${f_{{\rm{net}}}}$ toward centre
${\rm{T}} – {\rm{qE}} = {\rm{m}}\omega _{{\rm{particle }}}^2{\rm{R}}\quad \left[ {{{\rm{w}}_{{\rm{partice }}}} = {\theta _{\max }}{\omega _{{\rm{phase}}{\rm{. }}}}} \right]$
$\mathrm{T}=\mathrm{qE}+\mathrm{m} \frac{3 \mathrm{qE} \theta_{0}^{2}}{2 \mathrm{m} \ell}\left(\frac{2 \ell}{3}\right) \quad\left[\mathrm{R}=\frac{2 \ell}{3}\right]$
$\mathrm{T}=\mathrm{q} \mathrm{E}+\mathrm{q} \mathrm{E} \theta_{0}^{2}$
