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MHT CET 2024 (11 May, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from MHT CET 2024 (11 May, Shift 2)

Q7MHT CET 2024 · 11 MAY · SHIFT 2Easy

Motion of Charges in an Electric Field

A particle ' $A$ ' has charge ' $+q$ ' and a particle ' $B$ ' has charge ' $+4q$ '. Each has same mass ' $m$ '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)

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Answer: (b)

Energy gained: $\frac{1}{2}mv^2 = qV \Rightarrow v \propto \sqrt{q}$ for the same $m$ and $V$ .
$\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \frac{1}{2}$
So the answer is (b) $1:2$ .

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