Question

A rod AB of mass $m$ and length $l$ is positively charged with linear charge density $\lambda$ C/m. It is pivoted at end A and is hanging freely. If a horizontal electric field $E$ is switched on in the region, find the angular acceleration of the rod with which it starts

• A. $\frac{E\lambda }{2m}$
• B. $\frac{3E\lambda }{2m}$
• C. $\frac{3E\lambda }{m}$
• D. Zero

Answer 1

Answer: (B)

$\frac{3E\lambda }{2m}$

Torque for element $dx$ of the system, $d\tau =xdF=xE\lambda dx$ (as\ \ $dF=Eq=E\lambda dx$)
$\tau =E\lambda {\int }_0^{l}xdx=\frac{1}{2}E\lambda l^2$
Again, torque$\left(\tau \right)=$moment of inertia $\left(I\right)\times$ angular acceleration$\left(\alpha \right)$
So, $\frac{1}{2}E\lambda l^2=\frac{M{l}^2}{3}\times \alpha$
$\alpha =\frac{3E\lambda }{2M}$