A) \[r\]
B) \[2r\]
C) \[r/2\]
D) \[r/4\]
Correct Answer: D
Solution :
Let a particle of charge q having velocity\[v\]approaches Q upto a closest distance r and if the velocity becomes\[2v,\]the closest distance will be\[r\]. The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. Or \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{r}\] Or \[\frac{1}{2}m{{v}^{2}}=k\frac{Qq}{r}\] ...(i) \[\left( k=constant=\frac{1}{4\pi {{\varepsilon }_{0}}} \right)\] and \[\frac{1}{2}m{{(2v)}^{2}}=k\frac{Qq}{r}\] ?(ii) Dividing Eq. (i) by Eq. (ii), \[\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{\frac{kQq}{r}}{\frac{kQq}{r}}\] \[\Rightarrow \] \[\frac{1}{4}=\frac{r}{r}\] \[\Rightarrow \] \[r=\frac{r}{4}\]
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