"A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 105 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 :"

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Accepted Answer

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${v^2} - {u^2} = 2as$

$\Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} \right)(S)$

$$ \Rightarrow - {200^2} = 2\left[ {{{ - 40 \times {{10}^{ - 6}} \times {{10}^5}} \over {100 \times {{10}^{ - 6}}}}} \right][S]$$

$\Rightarrow S = {4 \over {2 \times 4}}$ m = 0.5 m

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A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 10⁵ 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 :

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A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 105 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 :

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A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 10⁵ 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 :