CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equipartition Theorem

The potential...

Question

The potential energy of a particle in a certain field has the form U = $\frac{a}{r^{2}}$ - $\frac{b}{r}$ , where a and b are postive constants , r is the distance from the centre of the field . Find the value of $r_{0}$ for which the paricle is in stable equilibrium

A

$r_{0}$ = $\frac{2 a}{b}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

$r_{0}$ = $\frac{a}{b}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

$r_{0}$ = $\frac{2 a}{b}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

$r_{0}$ = - $\frac{a}{2 b}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$r_{0}$ = $\frac{2 a}{b}$

U(r) = $\frac{a}{r^{2}}$ - $\frac{b}{r}$

Force = F = - $\frac{d u}{d r}$ = - $(\frac{- 2 a}{r^{3}} + \frac{b}{r^{2}}$ ) $\Rightarrow$ F = $\frac{(b r - 2 a)}{r^{3}}$

At equilibrium F = - $\frac{d u}{d r}$ = 0

hence br -2a = 0 at equilibrium

r = $r_{0}$ = $\frac{2 a}{b}$ corresponds to equilibrium.

At stable equilibrium the potential energy is minimum and at unstable equilibirum it is maxium . From calculating , we know that for minimum value around a point r = $r_{0}$ , the first derivative should be zero and the second derivative should be positive

For minimim potential energy , $\frac{d u}{d r}$ = 0 and $\frac{d u^{2}}{d r^{2}}$ = $\frac{d}{d r}$ ( $\frac{- 2 a}{r^{3}}$ + $\frac{b}{r^{2}}$ ) = $\frac{6 a}{r^{4}}$ - $\frac{2 b}{r^{3}}$

At r = $r_{0}$ = $\frac{2 a}{b}$ $\Rightarrow$ $\frac{d u^{2}}{d r^{2}}$ = $\frac{6 a - 2 b r_{0}}{{r_{0}}^{4}}$ = $\frac{2 a}{{r_{0}}^{4}}$ $>$ 0

Hence the potential energy funtio U has a minimum value at $r_{0}$ = $\frac{2 a}{b}$ . The system has a stable equilibrium at minimum potential energy state

Suggest Corrections

thumbs-up

0

Similar questions

Q. The potential energy of a particle in a certain field has the form

U = \frac{a}{r^{2}} - \frac{b}{r},

a

b

are postive constants,

r

is the distance from the centre of the field.

Q. The potential energy of a particle in a certain field has the form

U = \frac{a}{r^{2}} - \frac{b}{r},

a

b

are postive constants,

r

is the distance from the centre of the field.

Q. Two charged particles A and B repel each other by a force

\frac{k}{r^{2}}

, where k is a constant and r is the separation between them. The particle A is clamped to a fixed point in the lab and the particle B which has a mass m, is released from rest with an initial separation

r_{0}

from A. Find the change in the potential energy of the two particle system as the separation increases to a large value. What will be the speed of the particle B in this situation?

Q. The potential energy of a particle in a certain field is given by

U = \frac{a}{r^{2}} - \frac{b}{r},

where a and b are positive constants and r is the distance from the centre of the field. The distance of particle in the stable equilibrium position is

The wavefunction of a 2s electron is given by $ψ_{2 s} = \frac{1}{4 \sqrt{2} π} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{a_{0}}}$
It has a node at $r = r_{0}$ . The value of $\frac{r_{0}}{a_{0}}$ is___

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Theory of Equipartitions of Energy

PHYSICS

Watch in App

explore_more_icon

Explore more

Equipartition Theorem

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon