91Ó°ÊÓ

There is a point charge $q$.

There is a polarization charge surrounding $q$.

There is a surface charge${\mathrm{σ}}_b$ on the upper surface of the lower dielectric.

There is a surface charge $\mathrm{σ}{\text{'}}_b$ on the lower surface of the upper dielectric.

The susceptibility of the medium is ${\mathrm{χ}}_e$ and ${{\mathrm{χ}}^{\text{'}}}_e$.

The polarization charge due to $q$ is

${\mathit{q}}_{\mathbf{p}}\mathbf{=}\mathbf{-}\mathit{q}\frac{{{\mathbf{χ}}^{\mathbf{\text{'}}}}_{\mathbf{e}}}{\mathbf{1}\mathbf{+}{{\mathbf{χ}}^{\mathbf{\text{'}}}}_{\mathbf{e}}}$

The expressions for the surface bound charge densities are

$\begin{array}{l}{\mathrm{σ}}_b={\mathrm{Î\mu }}_0{\mathrm{χ}}_e\left[−\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qd}{{{\mathrm{Î\mu }}^{\text{'}}}_r{(r^2+d^2)}^{3/2}}−\frac{{\mathrm{σ}}_b}{2{\mathrm{Î\mu }}_0}−\frac{{{\mathrm{σ}}^{\text{'}}}_b}{2{\mathrm{Î\mu }}_0}\right]\\ {{\mathrm{σ}}^{\text{'}}}_b={\mathrm{Î\mu }}_0{{\mathrm{χ}}^{\text{'}}}_e\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qd}{{{\mathrm{Î\mu }}^{\text{'}}}_r{(r^2+d^2)}^{3/2}}−\frac{{\mathrm{σ}}_b}{2{\mathrm{Î\mu }}_0}−\frac{{{\mathrm{σ}}^{\text{'}}}_b}{2{\mathrm{Î\mu }}_0}\right]\end{array}$

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space and ${{\mathrm{Î\mu }}^{\text{'}}}_r$is the permittivity of the upper medium.

Solve the above equations to get

$\begin{array}{l}{\mathrm{σ}}_b=−\frac{1}{4\mathrm{Ï€}}\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{\mathrm{χ}}_e}{1+\frac{{\mathrm{χ}}_e+{{\mathrm{χ}}^{\text{'}}}_e}{2}}\\ {{\mathrm{σ}}^{\text{'}}}_b=\frac{1}{4\mathrm{Ï€}}\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{\mathrm{Î\mu }}_r{{\mathrm{χ}}^{\text{'}}}_e/{{\mathrm{Î\mu }}^{\text{'}}}_r}{1+\frac{{\mathrm{χ}}_e+{{\mathrm{χ}}^{\text{'}}}_e}{2}}\end{array}$

Here, ${\mathrm{Î\mu }}_r$is the permittivity of the lower medium.

The total bound surface charge is then

$\mathrm{σ}=\frac{1}{4\mathrm{Ï€}}\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{{\mathrm{χ}}^{\text{'}}}_e−{\mathrm{χ}}_e}{{{\mathrm{Î\mu }}^{\text{'}}}_r\left(1+\frac{{\mathrm{χ}}_e+{{\mathrm{χ}}^{\text{'}}}_e}{2}\right)}$

The total bound charge from the surface charge density is

$q_b=q\frac{{{\mathrm{χ}}^{\text{'}}}_e−{\mathrm{χ}}_e}{2{{\mathrm{Î\mu }}^{\text{'}}}_r\left(1+\frac{{\mathrm{χ}}_e+{{\mathrm{χ}}^{\text{'}}}_e}{2}\right)}$

The potential is thus

$V=\left[\begin{array}{l}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left\{\frac{q/{{\mathrm{Î\mu }}^{\text{'}}}_r}{\sqrt{x^2+y^2+{(z−d)}^2}}+\frac{q_b}{\sqrt{x^2+y^2+{(z+d)}^2}}\right\}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}z>0\\ \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q/({{\mathrm{Î\mu }}^{\text{'}}}_r+{\mathrm{Î\mu }}_r)}{\sqrt{x^2+y^2+{(z−d)}^2}}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰}z<0\end{array}\right]$

Thus, this is the expression for the potential everywhere.