Chapter 11, Problem 11.1E

Convert $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ into units of $\text{N}/\text{m}$.

Interpretation Introduction

Interpretation:

The value $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ is to be converted into units of $\text{N}/\text{m}$.

Concept introduction:

The Hook’s law states that the strain produced in a solid is directly proportional to the applied force. The classical harmonic oscillator shows repetitive motion. It follows Hook’s law. The Hook’s law is given by,

$F=-Kx$

Where,

• $F$ is the force.

• $K$ is the spring constant.

• $x$ is the deformation.

Answer to Problem 11.1E

The value of $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ in units of $\text{N}/\text{m}$ is $355.8\text{N}{\text{m}}^{-1}$.

Explanation of Solution

The conversion factor of $\text{dyn}$ into Newton is given by,

$\text{1}\text{dyn}={10}^{-5}\text{N}$

The conversion factor of m to nm is given by,

$\text{1}\stackrel{\circ }{\text{A}}={10}^{-10}\text{m}$

The conversion of $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ into $\text{N}/\text{m}$ is as follows:

$\begin{array}{rll}\frac{3.558\text{m}\text{dyn}}{\stackrel{\circ }{\text{A}}}& =& \frac{3.558\text{m}\text{dyn}\left(\frac{1\text{dyn}}{{10}^3\text{m}\text{dyn}}\right)\left(\frac{{10}^{-5}\text{N}}{1\text{dyn}}\right)}{\stackrel{\circ }{\text{A}}\times \left(\frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}\\ & =& 355.8{\text{Nm}}^{-1}\end{array}$

Hence, the value of $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ in units of $\text{N}/\text{m}$ is $355.8\text{N}{\text{m}}^{-1}$.

Conclusion

The value of $3.558\text{mdyn}/\stackrel{\circ }{\text{A}}$ in units of $\text{N}/\text{m}$ is $355.8\text{N}{\text{m}}^{-1}$.