Q. A particle A has charge $+ q$ and particle B has charge $+ 4q$ with each of them having the save mass $m.$ when allowed to fall from rest through the same electrical potential difference, the ration of their steeds $v_{A}/v_{B}$ will become
A) 2 : 1
B) 1 : 2
C) 1 : 4
D) 4 : 1
Explanation
(b) $\frac{E_{A}}{E_{B}} = \frac{\frac{1}{2}\ {mv}_{A}^{2}}{\frac{1}{2}\ {mv}_{B}^{2}} = \frac{W_{A}}{W_{B}} = \frac{(q)V}{(4q)V}$
$\frac{v_{A}}{v_{B}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.