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Equipartition Theorem

The potential...

Question

The potential energy of a particle in a certain field has the form U = $\frac{a}{r^{2}}$ - $\frac{b}{r}$ , where a and b are postive constants , r is the distance from the centre of the field . Find the value of $r_{0}$ for which the paricle is in stable equilibrium

A

$r_{0}$ = $\frac{2 a}{b}$

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B

$r_{0}$ = $\frac{a}{b}$

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C

$r_{0}$ = $\frac{2 a}{b}$

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D

$r_{0}$ = - $\frac{a}{2 b}$

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Solution

The correct option is A
$r_{0}$ = $\frac{2 a}{b}$

U(r) = $\frac{a}{r^{2}}$ - $\frac{b}{r}$

Force = F = - $\frac{d u}{d r}$ = - $(\frac{- 2 a}{r^{3}} + \frac{b}{r^{2}}$ ) $\Rightarrow$ F = $\frac{(b r - 2 a)}{r^{3}}$

At equilibrium F = - $\frac{d u}{d r}$ = 0

hence br -2a = 0 at equilibrium

r = $r_{0}$ = $\frac{2 a}{b}$ corresponds to equilibrium.

At stable equilibrium the potential energy is minimum and at unstable equilibirum it is maxium . From calculating , we know that for minimum value around a point r = $r_{0}$ , the first derivative should be zero and the second derivative should be positive

For minimim potential energy , $\frac{d u}{d r}$ = 0 and $\frac{d u^{2}}{d r^{2}}$ = $\frac{d}{d r}$ ( $\frac{- 2 a}{r^{3}}$ + $\frac{b}{r^{2}}$ ) = $\frac{6 a}{r^{4}}$ - $\frac{2 b}{r^{3}}$

At r = $r_{0}$ = $\frac{2 a}{b}$ $\Rightarrow$ $\frac{d u^{2}}{d r^{2}}$ = $\frac{6 a - 2 b r_{0}}{{r_{0}}^{4}}$ = $\frac{2 a}{{r_{0}}^{4}}$ $>$ 0

Hence the potential energy funtio U has a minimum value at $r_{0}$ = $\frac{2 a}{b}$ . The system has a stable equilibrium at minimum potential energy state

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