First slide

Equilibrium

Question

The potential energy of a particle in 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 is

Moderate

A

A/B
B

B/A
C

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

2A/B

Solution

$f o r s t a b l e e q u i l i b r i u m F = - \frac{d U}{d r} = 0 - \frac{d}{d r} (\frac{A}{r^{2}} - \frac{B}{r}) = 0 A (- 2 r^{- 3}) - B (- 1 r^{- 2}) = 0 \Rightarrow r = \frac{2 A}{B}$

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