Question:medium

An $\alpha$-particle of mass $6.4\times10^{-27}$ kg and charge $3.2\times10^{-19}$ C is situated in a uniform electric field of $1.6\times10^{5}$ V m$^{-1}$. The velocity of the particle at the end of $2\times10^{-2}$ m path when it starts from rest is

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Combine $F=qE$ and $v^2=2as$ for charged particles in electric fields.

Updated On: Apr 10, 2026

$2\sqrt{3}\times10^5~\text{ms}^{-1}$
$8\times10^5~\text{ms}^{-1}$
$16\times10^5~\text{ms}^{-1}$
$4\sqrt{2}\times10^5~\text{ms}^{-1}$

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The Correct Option is D

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