A particle located in a one-dimensional potential field has its potential energy function as `U(x)=(a)/(x^4)-(b)/(x^2)`, where a and b are positive constants. The position of equilibrium x corresponds to

Options

A. (a) `(b)/(2a)`
B. (b) `sqrt((2a)/(b))`
C. (c) `sqrt((2b)/(a))`
D. (d) `(a)/(2a)`

Answer

B

Explanation

The position of equilibrium corresponds to `F(x)=0`
Since `F(x)=(-dU(x))/(dx)`
so `F(x)=-(d)/(dx)(a/x^4-b/x^2)` or `F(x)=(4a)/(x^5)-(2b)/(x^3)`
For equilibrium, `F(x)=0`, therefore
`(4a)/(x^5)-(2b)/(x^3)=0impliesx=+-sqrt((2a)/(b))`
`(d^2U(x))/(dx^2)=-(20a)/(x^6)+(8b)/(x^4)`
Putting `x=+-sqrt((2a)/(6))` gives `(d^2U(x))/(dx^2)` as negative
So U is maximum. Hence, it is position of unstable equilibrium