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MHT CET 2021 (22 Sep, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from MHT CET 2021 (22 Sep, Shift 2)

Q25MHT CET 2021 · 22 SEP · SHIFT 2Easy

Motion of Charges in an Electric Field

Two particles $A$ and $B$ having same mass have charge $+q$ and $+4q$ respectively. When they are allowed to fall from rest through same electric potential difference, the ratio of their speeds ' $V_A$ ' to ' $V_B$ ' will become

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Answer: (a)

$\frac{1}{2}mv^2 = qV \Rightarrow v \propto \sqrt{q}$ for the same $m$ and $V$ .
$\frac{V_A}{V_B} = \sqrt{\frac{q}{4q}} = \frac{1}{2}$
So the answer is (a) $1:2$ .

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