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Q19E

Expert-verifiedFound in: Page 556

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**Suppose a force between two particles decreases distance according to ${F}{=}{k}{/}{{r}}^{{b}}$** **. What is the limit on b if the energy required to separate the particles ****Infinitely far is not to be infinite?**

The limit on b is $b>1$ .

A force between two particles decreases with distance, according to $F=k/{r}^{b}$ .

**The force ${{F}}_{{x}}$ is the negative of the derivative of the potential energy U such that,**

** ${{F}}_{{x}}{=}{-}\frac{dU}{dx}$**

**Therefore, the differential form of the potential energy can be written as, **

**${d}{U}{=}{-}{{F}}_{{x}}{d}{x}$**** **

Integrate the above equation on both sides, andgetthe expression for the total potential energy as:

$U\left(x\right)=-\int {F}_{x}dx$

For force,the formula can be written as:

$F=\frac{k}{{r}^{b}}$

The energy of this force would be given as,

$U=-\int Fdr$

Substitute $F=\frac{k}{{r}^{b}}$ in the above equation as:

$\begin{array}{rcl}U& =& \int \frac{k}{{r}^{b}}dr\\ U& =& -\int k{r}^{-b}dr\\ U& =& -\frac{k{r}^{-b+1}}{-b+1}\\ & & \end{array}$

When b is not 1 , we need $-b+1<0$ to make sure the energy is not infinite at infinite separation, which is $b>1$. Thus, the limit on is $b>1$

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