91Ó°ÊÓ

Write the expression for the Polarization is proportional to the electric field.

$P={\mathrm{Î\mu }}_0{\mathrm{χ}}_{e}E$ …… (1)

If the material consists of atom ( or nonpolar molecules ), the induced dipole moment of each one is likewise proportional to the field.

$p=\mathrm{Î\pm }E$ …... (2)

Here, is the dipole moment per unit volume and is the dipole moment per atom.

Write the relation between these two.

$P=Np$

$$\begin{gathered} P=N\left(\mathrm{Î\pm }E\right) \\ P=N\mathrm{Î\pm }E \end{gathered}$$

$\begin{array}{rcl}{\mathrm{Î\mu }}_0{\mathrm{χ}}_{e}E& =& N\mathrm{Î\pm }E\\ {\mathrm{χ}}_e& =& \frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\\ & & \end{array}$ ……. (3)

From the above equation (3) gives the relation between atomic polazabilityand susceptibility${\mathrm{χ}}_e$ and his equation is not valid if the density is very low.

Write the expression for the density of the atoms.

$N=\frac{1}{\left(\frac{4}{3}\right)\mathrm{Ï€}{R}^3}$

The macroscopic field E is given by,

$E=E{}_{self}+E_{else}$ …… (4)

Here,$E_{self}$is the average field over the sphere due to atom itself.

Write the expression for $E_{self}$.

$E_{self}=-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{p}{R^3}$ …… (5)

Write the expression for dipole moment per atom.

$p=\mathrm{Î\pm }{E}_{else}$

Write the expression for dipole moment per unit volume.

$\begin{array}{rcl}P& =& Np\\ & =& N\left(\mathrm{Î\pm }{E}_{else}\right)\\ & =& N\mathrm{Î\pm }{E}_{else}\\ & & \end{array}$ …… (6)

Thus,

Write the expression of macroscopic field.

$\begin{array}{rcl}E& =& E_{self}+E_{else}\\ & =& -\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{p}{R^3}+E_{else}\\ & =& -\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{Î\pm }{E}_{else}}{R^3}+E_{else}\\ & =& E_{else}\left[1-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{Î\pm }}{R^3}\right]\\ & & \end{array}$

Solve as further,

$\begin{array}{rcl}E& =& E_{else}\left[1-\frac{1}{4\mathrm{Ï€}{R}^3}\frac{\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\right]\\ & =& E_{else}\left[1-\frac{N}{3}\frac{\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\right]\left[âˆ\mu N=\frac{3}{4\mathrm{Ï€}{R}^3}\right]\\ & =& E_{else}\left[1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}\right]\\ & & \end{array}$

$E_{else}=\frac{E}{\left[1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}\right]}$

Now, substitute$\frac{E}{\left[1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}\right]}$for$E_{else}$in equation (6)

$\begin{array}{rcl}P& =& N\mathrm{Î\pm }\left(\frac{E}{\left[1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}\right]}\right)\\ & =& \left(\frac{N\mathrm{Î\pm }}{1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}}\right)E\\ & & \end{array}$ …… (7)

Now comparing equation (7) with equation (1),

$$\begin{gathered} {\mathrm{Î\mu }}_0{\mathrm{χ}}_e=\frac{N\mathrm{Î\pm }}{1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}} \\ {\mathrm{χ}}_e=\frac{\left(\frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\right)}{1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}} \\ {\mathrm{χ}}_e\left(1-\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}\right)=\frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0} \\ {\mathrm{χ}}_e-{\mathrm{χ}}_e\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}=\frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Solve as further,

$\begin{array}{rcl}{\mathrm{χ}}_e& =& {\mathrm{χ}}_e\frac{N\mathrm{Î\pm }}{3{\mathrm{Î\mu }}_0}+\frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\\ {\mathrm{χ}}_e& =& \frac{N\mathrm{Î\pm }}{{\mathrm{Î\mu }}_0}\left(1+\frac{{\mathrm{χ}}_e}{3}\right)\\ \mathrm{Î\pm }& =& \frac{{\mathrm{Î\mu }}_0}{N}\frac{{\mathrm{χ}}_e}{\left(1+\frac{{\mathrm{χ}}_e}{3}\right)}\\ \mathrm{Î\pm }& =& \frac{3{\mathrm{Î\mu }}_0}{N}\frac{{\mathrm{χ}}_e}{3+{\mathrm{χ}}_e}\\ & & \end{array}$.....(8)

But .${\mathrm{χ}}_e={∈}_r-1$

Substitute the value of ${\mathrm{χ}}_e$in equation (8)

$$\begin{gathered} \mathrm{Î\pm }=\frac{3{\mathrm{Î\mu }}_0}{N}\frac{\left({\mathrm{Î\mu }}_r-1\right)}{3+\left({\mathrm{Î\mu }}_r-1\right)} \\ \mathrm{Î\pm }=\frac{3{\mathrm{Î\mu }}_0}{N}\frac{\left({\mathrm{Î\mu }}_r-1\right)}{\left({\mathrm{Î\mu }}_r+2\right)} \end{gathered}$$