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:

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

The potential energy $(U)$ of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$,where $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,where $a = \dots \dots \dots \dots$

When a stone is thrown vertically upwards and returns to the ground,at what point will its potential energy be maximum?

The potential energy between two atoms is given by the function $U(r) = a/r^{12} - b/r^{6}$. Find the equilibrium distance between them.

What does a body at rest possess?