A uniform rod AB of mass m and length $l$ 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. Assume that electric field due to large non-conducting sheet is $\frac{σ}{2 ε_{0}}$ .Now the rod is rotated by a small angle in the plane of the paper and released. The rod performs SHM and its time period is $T = 2 π \sqrt{\frac{x m \in_{0}}{y λ σ}}$ .find $x + y$

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Detailed Solution

Whose magnitude is $τ = F \frac{l}{2} \sin θ = \frac{F l}{2} θ,$ But $F = λ \frac{l}{2} \frac{σ}{2 \in_{0}} = \frac{λ l σ}{4 \in_{0}}$
$∴ τ = (\frac{λ l^{2} σ}{8 \in_{0}}) θ$
Now, since $τ$ is towards the mean position & $τ \propto θ$
$τ = I α h \frac{λ l^{2} σ}{8 \in_{0}} θ \Rightarrow \frac{m l^{2}}{12} α = \frac{λ l^{2} σ}{8 \in_{0}} o r α = (\frac{3 λ σ}{2 m \in_{0}}) θ ∴ w^{2} = \frac{3 λ σ}{2 m \in_{0}} = {(\frac{2 π}{T})}^{2} \Rightarrow T = 2 π \sqrt{\frac{2 m \in_{0}}{3 λ σ}}$