First slide

Concept of potential energy

Question

The potential energy of a positive in a certain field is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ , where a and b are positive constant and r is the distance from centre of field. The equilibrium position of particle in this field is given by $r_{0} = \frac{n . a}{b}$ . The value of n is

Moderate

Solution

$\vec{F} = \frac{- d U}{d r}$
$\begin{array}{l} At equilibrium F = 0 \Rightarrow \frac{- d U}{d r} = 0 \\ - \frac{d}{d r} (\frac{a}{r^{2}} - \frac{b}{r}) = 0 \Rightarrow - \frac{- 2 a}{r^{3}} + \frac{b}{r^{2}} = 0 \\ \frac{b}{r^{2}} = \frac{2 a}{r^{3}} \\ r = \frac{2 a}{b} \Rightarrow n = 2 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App