Question

In the figure shown $S$ is a large non-conducting sheet of uniform charge density $\sigma$. A rod $R$ of length $\ell$ and mass ' $m$ ' is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. Find the angular acceleration of the rod just after it is released. If it is $\frac{a\sigma \lambda }{2m{\epsilon }_0}$. Find $a$.

Answer 1

Answer: 3.00

Electric field near the large nonconducting sheet
$E=\frac{\sigma }{2{\epsilon }_0}(\hat{n})$
Force $F$ on the hinged rod on upper & lower half each.
$F=\frac{\sigma \lambda (\ell )}{2{\epsilon }_0\times 2}$
Equal and opposite force on rod constitute a couple
So torque acting on rod $=$ force $\times \perp$ distance between force.
$\therefore \tau =\frac{\sigma \lambda \ell }{4{\epsilon }_0}\times \frac{\ell }{2}=\frac{\sigma \lambda {\ell }^2}{8{\epsilon }_0}$
$\tau =I\alpha$
moment of Inertia $I=\left(\frac{m{\ell }^2}{12}\right)$
$\therefore \alpha =\frac{\sigma \lambda {\ell }^2}{8{\epsilon }_0}\times \frac{12}{m{\ell }^2}=\frac{3}{2}\frac{\sigma \lambda }{{\epsilon }_{0}m}$
$\alpha =\frac{3\sigma \lambda }{2{\epsilon }_{0}m}$