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

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Q. 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 it will travel before coming to the rest momentarily :

JEE MainJEE Main 2022Electric Charges and Fields

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$