Home
Class 12
PHYSICS
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/2A

B

2A/B

C

A/B

D

B/A

Text Solution

AI Generated Solution

The correct Answer is:
To find the distance of the particle from the center of the force field for stable equilibrium, we start with the given potential energy function: \[ U = \frac{A}{r^2} - \frac{B}{r} \] where \( A \) and \( B \) are positive constants, and \( r \) is the distance from the center of the field. ### Step 1: Find the force from the potential energy The force \( F \) associated with a potential energy \( U \) is given by: \[ F = -\frac{dU}{dr} \] ### Step 2: Differentiate the potential energy We need to differentiate \( U \) with respect to \( r \): \[ U = \frac{A}{r^2} - \frac{B}{r} \] Differentiating \( U \): \[ \frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^2}\right) - \frac{d}{dr}\left(\frac{B}{r}\right) \] Using the power rule for differentiation: \[ \frac{d}{dr}\left(\frac{A}{r^2}\right) = -2A \cdot r^{-3} = -\frac{2A}{r^3} \] \[ \frac{d}{dr}\left(\frac{B}{r}\right) = -B \cdot r^{-2} = -\frac{B}{r^2} \] Thus, \[ \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} \] ### Step 3: Set the force to zero for stable equilibrium For stable equilibrium, the net force must be zero: \[ -\frac{dU}{dr} = 0 \] This leads to: \[ -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = 0 \] So we have: \[ \frac{2A}{r^3} = \frac{B}{r^2} \] ### Step 4: Solve for \( r \) Cross-multiplying gives: \[ 2A = Br \] Now, solving for \( r \): \[ r = \frac{2A}{B} \] ### Conclusion The distance of the particle from the center of the force field for stable equilibrium is: \[ r = \frac{2A}{B} \]
Promotional Banner

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 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

The potential energy of a particle varies with distance x from a fixed origin as U = (A sqrt(x))/( x^(2) + B) , where A and B are dimensional constants , then find the dimensional formula for AB .

The potential energy of a particle varies with distance x from a fixed origin as U = (A sqrt(x))/( x^(2) + B) , where A and B are dimensional constants , then find the dimensional formula for AB .

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 :-

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.