Question

A particle A has a charge +q and particle B has charge +4q with 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 $\frac{v_{A}}{v_{B}}$ will becomes

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

Answer 1

Answer: (B)

1 : 2

Answer 2

We know that kinetic energy $K = \frac{1}{2}mv^{2} = QV$. Since, m and V are same so, $v^{2} \propto Q$ ⇒ $\frac{v_{A}}{v_{B}} = \sqrt{\frac{Q_{A}}{Q_{B}}} = \sqrt{\frac{q}{4q}} = \frac{1}{2}.$