# The potential energy of particle in a force field is, ,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: Option 1) B / 2A Option 2) 2A / B Option 3) A / B Option 4) B / A

As we discussed @7996

for equilibrium

$\frac{du}{dr}=0 \Rightarrow \frac{-2A}{r^{3}}+\frac{B}{r^{2}}=0$

$\therefore r=\frac{2A}{B}$

for stable equilibrium

$\frac{d^{2}u}{dr^{2}}$  should be positive for the value of r

$\frac{d^{2}u}{dr^{2}} = \frac{6A}{r^{4}}-\frac{2B}{r^{3}}$ is positive value for

$r =\frac{2A}{{B}}$

Option 1)

B / 2A

This is incorrect option

Option 2)

2A / B

This is correct option

Option 3)

A / B

This is incorrect option

Option 4)

B / A

This is incorrect option

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