Question

The potential energy of a particle in a field is $U=\frac{a}{r^2}-\frac{b}{r}$, where a and b are constant. The value of r in terms of a and b where the force on the particle is zero will be:

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{2a}{b}$
• D. $\frac{2b}{a}$

Answer 1

The correct option is C $\frac{2a}{b}$

U=\int{-dr}
$U=\frac{a}{r^2}-\frac{b}{r}$
$\frac{-dU}{dr}=F$
$-F=\frac{-2a}{r^3}-\frac{-b}{r^2}$
$F=0=\frac{-2a}{r^3}-\frac{-b}{r^2}$
$0=\frac{1}{r^2}(\frac{2a}{r}-b)$
Two cases: 1/r=0 is invalid
$\frac{2a}{r}=b$
r=2a/b