A uniform rod AB of mass m and length l is hi | vedclass

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Question

(D) The sheet produces a uniform electric field $E = \frac{\sigma}{2\epsilon_0}$ directed away from the sheet.The force on the left half $(AC)$ is $F

Vedclass · Accepted Answer

$T = 2\pi \sqrt{\frac{2m\epsilon_0}{3\lambda\sigma}}$

Vedclass · Answer

(D) The sheet produces a uniform electric field $E = \frac{\sigma}{2\epsilon_0}$ directed away from the sheet.The force on the left half $(AC)$ is $F = qE = (\lambda \cdot \frac{l}{2}) \cdot \frac{\sigma}{2\epsilon_0} = \frac{\lambda l \sigma}{4\epsilon_0}$ acting to the left.The force on the right half $(CB)$ is $F = qE = (\lambda \cdot \frac{l}{2}) \cdot \frac{\sigma}{2\epsilon_0} = \frac{\lambda l \sigma}{4\epsilon_0}$ acting to the right.When the rod is rotated by a small angle $	heta$,the torque about the hinge $C$ is $	au = 2 \cdot F \cdot (\frac{l}{4} \cos 	heta) \approx 2 \cdot F \cdot \frac{l}{4} = \frac{Fl}{2} = \frac{\lambda l^2 \sigma}{8\epsilon_0}$.Since $	au = I\alpha$,where $I = \frac{ml^2}{12}$ is the moment of inertia about the center $C$,we have $\frac{ml^2}{12} \alpha = \frac{\lambda l^2 \sigma}{8\epsilon_0} 	heta$.$\alpha = \frac{12 \lambda \sigma}{8 m \epsilon_0} 	heta = \frac{3 \lambda \sigma}{2 m \epsilon_0} 	heta$.Comparing with $\alpha = \omega^2 	heta$,we get $\omega^2 = \frac{3 \lambda \sigma}{2 m \epsilon_0}$.The time period is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2m\epsilon_0}{3\lambda\sigma}}$.

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