A) B/A
B) B/2A
C) 2A/B
D) A/B
Correct Answer: C
Solution :
Force experienced by the particle in field, \[F=-\left( \frac{dU}{dr} \right)=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)\] At equilibrium \[F=0\,F=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)=0\] \[\Rightarrow \]\[r=\frac{2A}{B}\]You need to login to perform this action.
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