A) \[\frac{1}{m}\]
B) \[\frac{1}{\sqrt{m}}\]
C) \[\frac{1}{{{m}^{2}}}\]
D) m
Correct Answer: A
Solution :
[a] At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. |
\[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=\frac{KQq}{d}\Rightarrow d\propto \frac{1}{m}\] |
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