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A large sheet carries uniform surface charge density $σ$ . A rod of length 2l has a linear charge density $λ$ , on one half and - $λ$ , on the second half. The rod is hinged at the midpoint P and makes an angle $θ$ with the normal to the sheet. The torque experienced by the rod is

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$\frac{{σλl}^{2}}{2 ε_{0}} \sin θ$

$\frac{{σλl}^{2}}{ε_{0}} \sin θ$

$\frac{σλl}{2 ε_{0}}$

answer is B.

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Detailed Solution

$τ = (λlE) lsin θ$

$= {λl}^{2} \frac{σ}{2 ε_{0}} \sin θ = \frac{{σλl}^{2}}{2 ε_{0}} \sin θ$

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