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

Option: 1

1 \mathrm{~m}


Option: 2

5 \mathrm{~m}


Option: 3

10 \mathrm{~m}


Option: 4

0.5 \mathrm{~m}


Answers (1)

best_answer

\mathrm{a=\frac{-q E}{m}}

\mathrm{v^{2}=u^{2}+2as}

\mathrm{0=200^{2}+2 \times\left(\frac{-q E}{m}\right) \times s}

\mathrm{-200^{2}=\frac{-2 \times 40 \times 10^{-6} \times 10^{5}}{100 \times 10^{-6}} \times \mathrm{S}}

\mathrm{S=0.5 m}

Hence, the correct answer is Option (4)

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chirag

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