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Asked by adjacentcalliber10 | 18 May, 2019, 13:17: PM
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Figure shows the particle of mass m and charge q  at one end of rod of length L . 
Other end of rod is hinged at O so that the rod rotates freely in the plane of figure.
 
Direction of electric field E is as shown in figure.

From the position as shown by angle θ that is made by the rod with field direction  to
the position OAB as shown in figure  which is parallel to field, the particle is displaced to a
distance AB in the direction of electric field.  Workdone W by the particle for this displacement is given by
 
W = qE×AB = q E L (1 - cosθ)  .........................(1)  
 
This workdone will be equal to the kinetic energy of particle when the rod  is  parallelel to field.
 
Hence we have, (1/2) m v2 = q E L (1 - cosθ)  ........................(2)
 
where m is mass of particle and v is speed of particle when it reahes the position B.
 
hence speed of particle v is given by,  v = [ 2q E L (1 - cosθ) / m ]1/2
Answered by Thiyagarajan K | 19 May, 2019, 11:55: AM

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