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Asked by adjacentcalliber10
18th May 2019,
1:17 PM
Answered by Expert
Answer:
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 Expert
19th May 2019,
11:55 AM
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