The potential energy of a particle in a ...

The potential energy of a particle in a force field is `U=(A)/(r^(2))-(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/B

B/A

`(B)/(2A)`

2A/B

Text Solution

Verified by Experts

The correct Answer is:

Similar Questions

Explore conceptually related problems

The potential energy of a particle in a conservative field is U =a/r^3-b/r^2, where a and b are positive constants and r is the distance of particle from the centre of field. For equilibrium, the value of r is

The potential energy of a particle in a certain field has the form U=(a//r^2)-(b//r) , where a and b are positive constants and r is the distance from the centre of the field. Find the value of r_0 corresponding to the equilibrium position of the particle, examine whether this position is stable.

The potential energy function of a particle is given by U(r)=A/(2r^(2))-B/(3r) , where A and B are constant and r is the radial distance from the centre of the force. Choose the correct option (s)

The potential energy between two atoms in a molecule is given by U(x)= (a)/(x^(12))-(b)/(x^(6)) , where a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when

The potential energy between two atoms in a molecule is given by U=ax^(2)-bx^(2) where a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when x is equal to :-

The potential energy of a particle varies with distance x from a fixed origin as V= (Asqrt(X))/(X + B) where A and B are constants . The dimension of AB are

In a conservative force field we can find the radial component of force from the potential energy function by using F = -(dU)/(dr) . Here, a positive force means repulsion and a negative force means attraction. From the given potential energy function U(r ) we can find the equilibrium position where force is zero. We can also find the ionisation energy which is the work done to move the particle from a certain position to infinity. Let us consider a case in which a particle is bound to a certain point at a distance r from the centre of the force. The potential energy of the particle is : U(r )=(A)/(r^(2))-(B)/(r ) where r is the distance from the centre of the force and A andB are positive constants. Answer the following questions. If the total energy of the particle is E=-(3B^(2))/(16A) , and it is known that the motion is radial only then the velocity is zero at

The potential energy of a conservative force field is given by U=ax^(2)-bx where, a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral.

In a conservative force field we can find the radial component of force from the potential energy function by using F = -(dU)/(dr) . Here, a positive force means repulsion and a negative force means attraction. From the given potential energy function U(r ) we can find the equilibrium position where force is zero. We can also find the ionisation energy which is the work done to move the particle from a certain position to infinity. Let us consider a case in which a particle is bound to a certain point at a distance r from the centre of the force. The potential energy of the particle is : U(r )=(A)/(r^(2))-(B)/(r ) where r is the distance from the centre of the force and A andB are positive constants. Answer the following questions. The work required to move the particle from equilibrium distance to infinity is

The potential energy of a particle in a force field is `U=(A)/(r^(2))-(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

Text Solution

Similar Questions

Recommended Questions