| A charged particle q is shot towards another charged particle\[Q\]which is fixed, with a speed v. It approaches\[Q\]upto a closest distance r and then returns. If q was given a speed 2 v, the closest distance of approach would be [AIEEE 2004] |
 |
A) r
B) 2r
C) r/2
D) r/4
Correct Answer: D
Solution :
| [d] At the distance of closest approach the kinetic energy is just convert into potential energy. In other words we can say that the kinetic energy equals the potential energy of charge particle. |
| 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 wilt 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) |
| \[\left( k=cons\tan t=\frac{1}{4\pi {{\varepsilon }_{0}}} \right)\] |
| and \[\frac{1}{2}m{{(2v)}^{2}}=k\frac{Qq}{r'}\] ...(ii) |
| Dividing Eq. (i) by Eq. (ii). we get |
| \[\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{\frac{kQq}{r}}{\frac{kQq}{r'}}\] |
| \[\Rightarrow \]\[\frac{1}{4}=\frac{r'}{r}\] \[\Rightarrow \]\[r'=\frac{r}{4}\] |
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