The potential energy of a particle in a certain field has the form U=ar2−br, where a and b are positive constants, r is the distance from centre of the field. At r=N1ab, particle is in equilibrium. Magnitude of maximum force of attraction is b3N2a2. Find the product of N1,N2

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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 positive constants, r is the distance from centre of the field. At $r = \frac{N_{1} a}{b},$ particle is in equilibrium. Magnitude of maximum force of attraction is $\frac{b^{3}}{N_{2} a^{2}} .$ Find the product of $N_{1}, N_{2}$

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Detailed Solution

$F = - \frac{d u}{d r} = - [\frac{- 2 a}{r^{3}} + \frac{b}{r^{2}}] For F=0, we get r= \frac{2a}{b} . \Rightarrow N_{1} =2$
With this r, $\frac{d^{2} u}{d r^{2}} = \frac{b}{r^{2}}$ , positive, so, u is minimum or stable equilibrium.
If F is maximum, $\frac{d F}{d r} = 0 \Rightarrow r = \frac{3 a}{b}$
With this r, we get $F = \frac{b^{3}}{27 a^{2}} \Rightarrow N_{2} = 27$