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

• A. $2:1$
• B. $1:2$
• C. $1:4$
• D. $4:1$

Answer 1

Answer: (B)

$1:2$

Answer 2

Using $v = \sqrt{\frac{2QV}{m}}$ ⇒ $v \propto \sqrt{Q}$⇒ $\frac{v_{A}}{v_{B}} = \sqrt{\frac{Q_{A}}{Q_{B}}} = \sqrt{\frac{q}{4q}} = \frac{1}{2}$