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Question

Physics Question on work, energy and power

The potential energy of a particle in a force field is U=Ar2Br 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

A

B2A\frac{B}{2A}

B

2AB\frac{2A}{B}

C

AB\frac{A}{B}

D

BA\frac{B}{A}

Answer

2AB\frac{2A}{B}

Explanation

Solution

For Stable equilibrium, F=dUdr=0F =-\frac{ d U }{ dr}=0
we get r=2A/Br =2 A / B