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A positive charged particle of 100mg is thrown in opposite direction to a uniform electric field of strength $10^{5} N / C$ . If the charge on the particle is $40 μ C$ and the initial velocity is 200m/s, how much distance it will travel before coming to the rest momentarily.

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0.5m

10m

answer is D.

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Detailed Solution

$a = \frac{F}{m} = \frac{E q}{m} = \frac{10^{5} \times 40 \times 10^{- 6}}{100 \times 10^{- 3} \times 10^{- 3}} = \frac{40 \times 100}{100 \times 10^{- 3}} = 40 \times 10^{3}$

$V^{2} - u^{2} = 2 a s$

$0^{2} - {(200)}^{2} = 2 \times 40 \times 10^{3} \times s$

$∴ S = \frac{4 \times 10^{4}}{2 \times 40 \times 10^{3}} = \frac{1}{2} = 0.5 m$

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