A uniform rod AB of mass m and length λ is hinged at its mid point C. The left half (AC) of the rod has linear charge density –λ and the right half (CB) has + λ, where λ is constant. A large non conducting sheet of uniform surface charge density σ is also present near the rod. Initially, the rod is kept perpendicular to the sheet. The end A of the rod is initially at a distance d. Now the rod is rotated by a small angle in the plane of the paper and released. Prove that the rod will perform SHM and find its time period.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
T = 2 π 
Sol. The sheet produces a uniform electric field E = 
towards right. The part AC and CB will experience electric force F as shown. They can be considered to be acting at the mid points of those parts respectively. The rod will experience torque about the point ‘c’ in the anticlockwise direction
Whose magnitude is τ = F 
sin θ 
θ ; But F = λ. 
. 
= 
∴ = 
θ
Now, since τ is towards the mean position & τ ∝ θ
∴ it will perform SHM → Hence proved
& τ = I α = 
⇒ 
= 
or α = 
∴ ω 2 = 
= 
T = 2π 
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