Answer to Question #86338 in Electricity and Magnetism for Rimjhim

Question

The polarisability of a given material is 1.0×10-40 C2
mN−1
. Determine its dielectric
constant if the number of molecules in 1 m3
volume of the material is 5.0×1025.

Answer 1

Given values, expressed in the basic SI units:

• polarisability: $\alpha = 1.0*10^{-40}~\text{kg}^{-1}\text{s}^{4}\text{A}^2;$
• concentration of molecules: $C = 5.0*10^{25}~\text{m}^{-3}.$

We will also need the vacuum permittivity (a.k.a. electric constant) for finding the relative permittivity (a.k.a. dielectric constant):

$\epsilon_0 \approx 8.854*10^{-12}~\text{m}^{-3}\text{kg}^{-1}\text{s}^{4}\text{A}^{2}.$

Paying attention to the dimentions (units), the relative permittivity will be expressed simply as:

$\epsilon = \frac{1}{\epsilon_0}\alpha C = \frac{1.0*10^{-40}~\text{kg}^{-1}\text{s}^{4}\text{A}^2~*~5.0*10^{25}~\text{m}^{-3}}{8.854*10^{-12}~\text{m}^{-3}\text{kg}^{-1}\text{s}^{4}\text{A}^{2}} \approx 5.647*10^{-4}.$

For any confusion about the terminology of the discussed values of permittivity, see Wikipedia:

• https://en.wikipedia.org/wiki/Vacuum_permittivity
• https://en.wikipedia.org/wiki/Relative_permittivity