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Two particles, each of mass $m$ , are attached to the ends of a light rigid rod of length $l$ . The particles are given charges $- q$ and $+ q$ and placed in a uniform electric field of strength $E$ . If the rod is slightly displaced from its stable position, it performs angular $SHM$ of angular frequency :

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$\sqrt{\frac{q E}{m l}}$

$\sqrt{\frac{2 q E}{m l}}$

$\sqrt{\frac{q E}{2 m l}}$

$\sqrt{\frac{4 q E}{m l}}$

answer is A.

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Detailed Solution

The system is a dipole of strength $p = q l$ .

When it is at angle $θ$ from its stable position, it experiences a restoring torque

$T_{res} = p E \sin θ = q l E θ (for small θ)$

The instantaneous angular acceleration is

$α = \frac{- T_{res}}{I} = \frac{- q I E}{2 m {(\frac{l}{2})}^{2}} θ = \frac{- 2 q E}{m l} θ$

Hence, the dipole perform angular SHM with angular frequency

$ω = \sqrt{\frac{2 q E}{m l}}$ .

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