First slide

Potential energy

Question

The potential energy of a particle in a force filed 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

Difficult

A

A/B
B

B/A
C

$\frac{B}{2 A}$
D

$\frac{2 A}{B}$

Solution

For stable equilibrium $\frac{d u}{d r} = 0$
$\frac{- 2 A}{r^{3}} + \frac{B}{r^{2}} = 0$
r = $\frac{2 A}{B}$

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