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Q.

A point particle of mass m is attached to one end of massless rigid non-conducting rod of length l. Another point particle of same mass is attached to the other end of the rod. The two particles carry equal charges +q and -q respectively. This arrangement is held in a region with the field E such that the rod makes an angle θ with the field direction :

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a

The tension in rod remains constant

b

If q is very small, the rod executes simple harmonic motion of period 2πml2qE

c

At every instant, net force on the system is zero

d

Both (b) and ( c ) are correct

answer is D.

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Detailed Solution

τ=-pE sinθFor small displacement sinθ≈θIα=-pEθ α∝θThis shows that electric dipole shows angular S.H.MIn linear simple harmonic motion.ω=km→spring constant→mass of oscillatorIn angular S.H.M., ω=pEI⇒2πT=pEI∴I = M.I. of rod about the centre = ml22∵p=2ql⇒T=2πIpE ∴T=2πml2qE Tension in the rod depends on the orientation of the dipole

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A point particle of mass m is attached to one end of massless rigid non-conducting rod of length l. Another point particle of same mass is attached to the other end of the rod. The two particles carry equal charges +q and -q respectively. This arrangement is held in a region with the field E such that the rod makes an angle θ with the field direction :