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

 

Answers (1)
A Aadil Khan

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}


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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