Question

Suppose the region abovethe xyplane in Ex. 4.8 is alsofilled withlinear dielectric but of a different susceptibility ${{\chi }^{\text{'}}}_e$.Find the potential everywhere.

Short answer

The potential is $V=\left[\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}\left\{\frac{q/{{\epsilon }^{\text{'}}}_r}{\sqrt{x^2+y^2+{(z-d)}^2}}+\frac{q_b}{\sqrt{x^2+y^2+{(z+d)}^2}}\right\}z>0\\ \frac{1}{4\pi {\epsilon }_0}\frac{2q/({{\epsilon }^{\text{'}}}_r+{\epsilon }_r)}{\sqrt{x^2+y^2+{(z-d)}^2}}z<0\end{array}\right]$

Step-by-step solution

Step 1:Given data

There is a point charge $q$.

There is a polarization charge surrounding $q$.

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

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

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

Define the polarization charge

The polarization charge due to $q$ is

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

Derive the expression for the potential

The expressions for the surface bound charge densities are

$\begin{array}{l}{\sigma }_b={\epsilon }_0{\chi }_e\left[-\frac{1}{4\pi {\epsilon }_0}\frac{qd}{{{\epsilon }^{\text{'}}}_r{(r^2+d^2)}^{3/2}}-\frac{{\sigma }_b}{2{\epsilon }_0}-\frac{{{\sigma }^{\text{'}}}_b}{2{\epsilon }_0}\right]\\ {{\sigma }^{\text{'}}}_b={\epsilon }_0{{\chi }^{\text{'}}}_e\left[\frac{1}{4\pi {\epsilon }_0}\frac{qd}{{{\epsilon }^{\text{'}}}_r{(r^2+d^2)}^{3/2}}-\frac{{\sigma }_b}{2{\epsilon }_0}-\frac{{{\sigma }^{\text{'}}}_b}{2{\epsilon }_0}\right]\end{array}$

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

Solve the above equations to get

$\begin{array}{l}{\sigma }_b=-\frac{1}{4\pi }\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{\chi }_e}{1+\frac{{\chi }_e+{{\chi }^{\text{'}}}_e}{2}}\\ {{\sigma }^{\text{'}}}_b=\frac{1}{4\pi }\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{\epsilon }_r{{\chi }^{\text{'}}}_e/{{\epsilon }^{\text{'}}}_r}{1+\frac{{\chi }_e+{{\chi }^{\text{'}}}_e}{2}}\end{array}$

Here, ${\epsilon }_r$is the permittivity of the lower medium.

The total bound surface charge is then

$\sigma =\frac{1}{4\pi }\frac{qd}{{(r^2+d^2)}^{3/2}}\frac{{{\chi }^{\text{'}}}_e-{\chi }_e}{{{\epsilon }^{\text{'}}}_r\left(1+\frac{{\chi }_e+{{\chi }^{\text{'}}}_e}{2}\right)}$

The total bound charge from the surface charge density is

$q_b=q\frac{{{\chi }^{\text{'}}}_e-{\chi }_e}{2{{\epsilon }^{\text{'}}}_r\left(1+\frac{{\chi }_e+{{\chi }^{\text{'}}}_e}{2}\right)}$

The potential is thus

$V=\left[\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}\left\{\frac{q/{{\epsilon }^{\text{'}}}_r}{\sqrt{x^2+y^2+{(z-d)}^2}}+\frac{q_b}{\sqrt{x^2+y^2+{(z+d)}^2}}\right\}z>0\\ \frac{1}{4\pi {\epsilon }_0}\frac{2q/({{\epsilon }^{\text{'}}}_r+{\epsilon }_r)}{\sqrt{x^2+y^2+{(z-d)}^2}}z<0\end{array}\right]$

Thus, this is the expression for the potential everywhere.