91Ó°ÊÓ

A spherical conductor, of radius a,carries a charge Q.

The spherical conductor is surrounded by a linear dielectric material of susceptibility $Xe$out to radius b.

The electrostatic energy in a configuration is given by

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{∫}\stackrel{\mathbf{→}}{\mathbf{D}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{E}}\mathit{d}\mathit{r}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{∫}\stackrel{\mathbf{→}}{\mathbf{D}}\mathbf{.}\mathit{E}\mathit{d}\mathit{r}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\stackrel{→}{D}$is the displacement current,$\stackrel{→}{E}$is the electric field and$dr$is the infinitesimal volume element.

The infinitesimal volume element for radial symmetry is

$\mathbf{»åÏ„}\mathbf{=}\mathbf{4}{\mathbf{Ï„Ï„°ù}}^{\mathbf{2}}\mathbf{dr}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, $r$is the radial coordinate.

The displacement current of the configuration is

$\stackrel{→}{D}=\left\{\begin{array}{ll}0& r<a\\ \frac{Q}{4{\mathrm{Ï€°ù}}^2}& r>a\end{array}\right.$

The electric field of the configuration is

$\stackrel{→}{E}\left\{\begin{array}{ll}\begin{array}{cc}0& r<a\\ \frac{Q}{4{\mathrm{Ï€Î\mu °ù}}^2}& a<r<b\\ \frac{Q}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}& r>b\end{array}& \end{array}\right.$

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space and $\mathrm{Î\mu }$is the permittivity of the surrounding

Thus, using equations (1) and (2), the energy of the configuration is

$$\begin{gathered} E=\frac{1}{2}\frac{Q}{{\left(4\mathrm{Ï€}\right)}^2}4\mathrm{Ï€}\left\{\frac{1}{3}{∫}_{\mathrm{a}}^{\mathrm{b}}\frac{1}{{\mathrm{r}}^2}\mathrm{dr}+\frac{1}{{\mathrm{Î\mu }}_0}{∫}_{\mathrm{b}}^{∞}\frac{1}{2}\mathrm{dr}\right\} \\ =\frac{\mathrm{Q}}{8{\mathrm{Ï€Î\mu }}_0}\left\{\frac{1}{1+\mathrm{Xe}}{\left[-\frac{1}{\mathrm{r}}\right]}_{\mathrm{a}}^{\mathrm{b}}+{\left[-\frac{1}{\mathrm{r}}\right]}_{\mathrm{b}}^{∞}\right\} \\ =\frac{{\mathrm{Q}}^2}{8{\mathrm{Ï€Î\mu }}_0\left(1+\mathrm{Xe}\right)}\left(\frac{1}{\mathrm{a}}+\frac{\mathrm{Xe}}{\mathrm{b}}\right) \end{gathered}$$

Thus, the energy of the configuration is $\frac{{\mathrm{Q}}^2}{8{\mathrm{Ï€Î\mu }}_0\left(1+\mathrm{Xe}\right)}\left(\frac{1}{\mathrm{a}}+\frac{\mathrm{Xe}}{\mathrm{b}}\right)$.