Question

Two point masses m each carrying charge – q and + q are attached to the ends of a mass less rigid non-conducting rod of length ℓ. The arrangement is placed in a uniform electric field E such that the rod makes a small angle 5° with the field direction. The minimum time needed by the rod to align itself along the field is …….. 

(A) t = π√[(2Mℓ)/(3qE)] 

(B) t = (π/2)√[(Mℓ)/(2qE)] 

(C) t = √[(Mℓ/qE)] 

(D) t = 2π√[(Mℓ/E)]

Answer 1

The correct option (B) t = (π/2)√[(Mℓ)/(2qE)

Explanation:

It is an electric dipole.

τ = Force × perpendicular distance

= F × AC

= F × AB sinθ

= qE × ℓ × sinθ

as θ = 5° sin θ is very small

 ∴ τ = qEℓθ

τ  = Rotating torque = – qEℓθ (1)

∵ τ = Iα

& I = M × (AO)² + M(BO)²

= M(ℓ/2)² + M(ℓ/2)² = (Mℓ/2)²

∴ τ = (Mℓ²/2) α

i.e. α = (2τ/Mℓ²)

from (1)

α = [(– 2qEℓθ)/(Mℓ²)] = [(– 2qE)/(Mℓ)]θ

as α = – ω²θ

∴ ω² = [(2qE)/(Mℓ)]    (2)

Time period = T = (2π/ω)

T = [(2π)/{(2qE)/(Ml)}^(1/2)]  From (2)

Rotating in clockwise sense, the minimum time taken by rod to align itself parallel to field = time to complete (1/4)^th of oscillation

∴ t_min = (T/4) = (π/2) [(Mℓ)/(2qE)]^(1/2)