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An electric dipole is formed by two particles fixed at the ends of a light rigid rod of length l. The mass of each particle is m and charges are –q and +q. The system is suspended by a torsion less thread in an electric field of intensity E such that the dipole axis is parallel to the filed if it is slightly displaced, the period of angular motion is:

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$2 π \sqrt{\frac{ml}{4 qE}}$

$\frac{1}{2 π} \sqrt{\frac{2 qE}{ml}}$

$2 π \sqrt{\frac{ml}{qE}}$

$2 π \sqrt{\frac{ml}{2 qE}}$

answer is C.

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Detailed Solution

$τ = pEsinθ τ = I α = pEsinθ$

$α = \frac{p E}{I} θ (a s θ i s s m a l l)$

$I = \frac{m l^{2}}{2}$

$ω = \sqrt{\frac{2 p E}{m l^{2}}}$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m l^{2}}{2 p E}}$

$p = q l$

$T = 2 π \sqrt{\frac{m l}{2 q E}}$

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