First slide

Potential energy

Question

A particle A has charge +q and a particle B has charge +4q with each of them having the same mass m. When al.lowed to fall from rest through the same electric potential difference, the ratio of their speed $\frac{v_{A}}{v_{B}}$ will become

Easy

A

2 : 1
B

1 : 2
C

1 : 4
D

4 : 1

Solution

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

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