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  4. A particle A has charge +q and particle B has charge +4q, each of them having the same mass m. When allowed to fall from rest through the same· electrical potential difference, the ratio of their speeds will become

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Q. A particle $A$ has charge $+q$ and particle $B$ has charge $+4q$, each of them having the same mass $m$. When allowed to fall from rest through the same· electrical potential difference, the ratio of their speeds will become

COMEDKCOMEDK 2015Electrostatic Potential and Capacitance

Solution:

Mass of each charged particle = $m$
Let potential difference be $V$.
The energy of charge $+q$ when passing through potential difference $V$,
$E = qV = \frac{1}{2} mv^2$
The energy of charge $+4q$ when passing through potential difference $V$,
, $E ' = 4q \, V = \frac{1}{2} mv'^2$
$\therefore \:\:\: \frac{E}{E'} =\frac{v^{2}}{v'^{2}} = \frac{q V}{4q V} =\frac{1}{4} $ or $\frac{v}{v'} =\frac{1}{2}$