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Question

# The potential energy of a particle in a certain field has the form U = ar2 - br, where a and b are postive constants , r is the distance from the centre of the field . Find the value of r0 for which the paricle is in stable equilibrium

A

r0 = 2ab

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B

r0 = ab

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C

r0 = 2ab

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D

r0 = - a2b

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Solution

## The correct option is A r0 = 2ab U(r) = ar2 - br Force = F = - dudr = - (−2ar3 +br2 )⇒ F = (br−2a)r3 At equilibrium F = - dudr = 0 hence br -2a = 0 at equilibrium r = r0 = 2ab 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 = r0, the first derivative should be zero and the second derivative should be positive For minimim potential energy , dudr = 0 and du2dr2 = ddr ( −2ar3 + br2) = 6ar4 - 2br3 At r = r0 = 2ab ⇒ du2dr2 = 6a−2br0r04 = 2ar04 > 0 Hence the potential energy funtio U has a minimum value at r0 = 2ab. The system has a stable equilibrium at minimum potential energy state

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