A positive charge particle of 100 ,mg is thrown in opposite direction to a uniform electric field of

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1. Electric Charges and Fields

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

A

$1$

B

$5$

C

$10$

D

$0.5$

(JEE MAIN-2022)

Solution

Distance travelled by particle before stopping

$\frac{ V ^{2}}{2 a }= S \Rightarrow \frac{ v ^{2} m }{2 qE }$

$\Rightarrow \frac{(200)^{2} \times 100 \times 10^{-6}}{2 \times 40 \times 10^{-6} \times 10^{5}}=0.5 \,m$

Standard 12

Physics

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