Q.

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×105NC1. If the charge on the particle is 40μC and the initial velocity is 200ms1, how much distance it will travel before coming to the rest momentarily :

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a

5 m

b

1 m

c

10 m

d

0.5 m

answer is D.

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Detailed Solution

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Distance travelled by particle before stopping
V22a=Sv2m2qE(200)2×100×1062×40×106×105=0.5m

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