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Question

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


A
Zero
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B
σλl28ϵ0sinθ
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C
σλl2ϵ0sinθ
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D
σλl22ϵ0
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Solution

The correct option is B σλl28ϵ0sinθ

We know that electric field due to infinite sheet is given as

E=σ2ϵ0

Charge on OB, q1=λl2

So, Force on q1,

F1=q1E=λσl4ϵ0 (towards right)

Further torque due to F1 about O,

τ1=λσl4ϵ0×l4sinθ

τ1=σλl216ϵ0sinθ (clockwise)

Similarly, for OA,

τ2=σλl216ϵ0sinθ (clockwise)

Further,
τnet=τ1+τ2

τnet=σλl28ϵ0sinθ (clockwise)

Hence, option (b) is correct answer.

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