A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 x 10^5 NC^-1. If the charge on the particle is 40µC and the initial velocity is 200 ms^-…

Question

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 x 10^5 NC^-1. If the charge on the particle is 40µC and the initial velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily:

Accepted Answer

Correct answer: D. (d) Distance travelled by particle before stopping V^22a=S ⇒ v^2m2 qE ⇒ (200 )^2 100 10^-62 40 10^-6 10^5=0.5m

$A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 x 1 0^{5} N C^{- 1} . If the charge on the particle is 40µC and the initial velocity is 200 m s^{- 1}, how much distance it will travel before coming to the rest momentarily:$

Solution

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