Electrostatics Question 359
Question: A particle $ A $ has charge $ +q $ and a particle $ B $ has charge $ +4q $ with each of them having the same mass $ m $ . When allowed to fall from rest through the same electric potential difference, the ratio of their speed $ \frac{v _{A}}{v _{B}} $ will become [BHU 1995; MNR 1991; UPSEAT 2000; Pb PET 2004]
Options:
A) $ 2:1 $
B) $ 1:2 $
C) $ 1:4 $
D) $ 4:1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Using $ v=\sqrt{\frac{2QV}{m}} $
therefore $ v\propto \sqrt{Q} $
therefore $ \frac{v _{A}}{v _{B}}=\sqrt{\frac{Q _{A}}{Q _{B}}}=\sqrt{\frac{q}{4q}}=\frac{1}{2} $
