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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
A
$\frac{B}{A}$
B
$\frac{B}{2A}$
C
$\frac{2A}{B}$
D
$\frac{A}{B}$
Detailed Solution
$F = -\frac{dU}{dr} = -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = \frac{2A}{r^3} - \frac{B}{r^2}$
For equilibrium F = 0: $\frac{2A}{r^3} = \frac{B}{r^2}$
$r = \frac{2A}{B}$
At this r, $\frac{d^2U}{dr^2} > 0$ (U is a minimum), so the equilibrium is stable.
For equilibrium F = 0: $\frac{2A}{r^3} = \frac{B}{r^2}$
$r = \frac{2A}{B}$
At this r, $\frac{d^2U}{dr^2} > 0$ (U is a minimum), so the equilibrium is stable.