The potential energy of a particle in a force | vedclass

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Question

(B) Given potential energy: $U(r) = \frac{A}{r^2} - \frac{B}{r}$.For equilibrium,the force $F = -\frac{dU}{dr} = 0$,which implies $\frac{dU}{dr} = 0$.

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$\frac{2A}{B}$

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(B) Given potential energy: $U(r) = \frac{A}{r^2} - \frac{B}{r}$.For equilibrium,the force $F = -\frac{dU}{dr} = 0$,which implies $\frac{dU}{dr} = 0$.$\frac{dU}{dr} = \frac{d}{dr}(Ar^{-2} - Br^{-1}) = -2Ar^{-3} + Br^{-2} = 0$.$\frac{B}{r^2} = \frac{2A}{r^3} \implies r = \frac{2A}{B}$.For stable equilibrium,the second derivative must be positive: $\frac{d^2U}{dr^2} > 0$.$\frac{d^2U}{dr^2} = \frac{d}{dr}(-2Ar^{-3} + Br^{-2}) = 6Ar^{-4} - 2Br^{-3}$.Substituting $r = \frac{2A}{B}$:$\frac{d^2U}{dr^2} = 6A(\frac{B}{2A})^4 - 2B(\frac{B}{2A})^3 = 6A(\frac{B^4}{16A^4}) - 2B(\frac{B^3}{8A^3}) = \frac{3B^4}{8A^3} - \frac{B^4}{4A^3} = \frac{3B^4 - 2B^4}{8A^3} = \frac{B^4}{8A^3}$.Since $A, B > 0$,$\frac{B^4}{8A^3} > 0$. Thus,the condition for stable equilibrium is satisfied at $r = \frac{2A}{B}$.

The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

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