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 $40 μ C$ and the initial velocity is $200 {ms}^{- 1}$ , how much distance it will travel before coming to the rest momentarily :

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5 m

1 m

10 m

0.5 m

answer is D.

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Detailed Solution

Distance travelled by particle before stopping
$\frac{V^{2}}{2 a} = S \Rightarrow \frac{v^{2} m}{2 qE} \Rightarrow \frac{(200)^{2} \times 100 \times 10^{- 6}}{2 \times 40 \times 10^{- 6} \times 10^{5}} = 0.5 m$

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