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Question

A particle of mass m and charge q is attached to a light rod of length L. The rod can rotate freely in the plane of paper about the other end, which is hinged at P. The entire assembly lies in a uniform electric field E also acting in the plane of paper as shown. The rod is released from rest when it makes an angle θ with the electric field direction. (Assume gravity free space)



A
Direction of angular acceleration changes alternatively
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B
If θ is small, the particle performs SHM
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C
The speed of the particle when the rod is parallel to the electric field is (2qEL(1sinθ)m)12
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D
The speed of the particle when the rod is parallel to the electric field is(2qEL(1cosθ)m)12
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Solution

The correct option is D The speed of the particle when the rod is parallel to the electric field is(2qEL(1cosθ)m)12
Since electric field is rightwards A is at highest potential
VAVB=(E)(AB)

=El(1cosθ)=El(1cosθ)

UAUB=q(VAVB)

=12mv2 12mv20


V0=0
El(1cosθ)(q)=12mv2

[v=(2qEl(1cosθ)m)12

For small θ, oscillation is SHM.

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