Question

A particle located in one dimensional potential field has potential energy function \( U(x)=\frac{a}{x^{2}}-\frac{b}{x^{3}} \), where \( a \) and \( b \) are positive constants. The position of equilibrium corresponds to \( x \) equal to

(1) \( \frac{3 a}{2 b} \) 

(2) \( \frac{2 b}{3 a} \) 

(3) \( \frac{2 a}{3 b} \) 

(4) \( \frac{3 b}{2 a} \)

Answer 1

\(U = \frac a{x^2} - \frac b{x^3}\)

At equilibrium Net force

\(F_{net} (x) = 0\)   .....(i)

\(F(x) = \frac d{dx} \left(\frac a{x^2} - \frac b{x^3}\right)\)

\(F(x) = -\frac{2a}{x^3} - \frac{(-3)b}{x^4}\)

\(F(x) = \frac{3b}{x^4} - \frac {2a}{x^3}\)

From (i),

\(\frac{3b}{x^4} - \frac{2a}{x^3} = 0\)

\(\frac{3b - 2ax}{x^4} = 0\)

\( 3{b} = 2ax\)

\(x = \frac{3b}{2a}\)