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 : 1
• B. 1 : 2
• C. 1 : 4
• D. 4 : 1

Answer 1

Answer: (B)

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