The potential energy of a particle in a force field is $U = \frac{A}{r^{2}} - \frac{B}{r}$, where $A$ and $B$ are positive constants and $r$ is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

Question

Dr. Abubacker · Accepted Answer

(b) For equilibrium
$\frac{dU}{dr} = 0\  \Rightarrow \frac{- 2A}{r^{3}} + \frac{B}{r^{2}} = 0$
$r = \frac{2A}{B}$
For stable equilibrium
$\frac{d^{2}U}{dr^{2}}$ should be positive for the value of $r$
Here $\frac{d^{2}U}{dr^{2}\ } = \frac{6A}{r^{4}} - \frac{2B}{r^{3}}$ is +ve value for $r = \frac{2A}{B}$

Q. The potential energy of a particle in a force field is $U = \frac{A}{r^{2}} - \frac{B}{r}$, where $A$ and $B$ are positive constants and $r$ is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

Explanation

Reviewed by Dr. Abubacker

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