Question

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

• A. 2:01
• B. 1:02
• C. 1:04
• D. 4:01

Answer 1

Answer: (B)

1:02

Answer 2

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 $\frac{v}{v'} =\frac{1}{2}$