Question

A spherical conductor, of radius a,carries a charge Q(Fig. 4.29). It

is surrounded by linear dielectric material of susceptibility$\mathit{X}\mathit{e}$out to radius b.Find the energy of this configuration (Eq. 4.58).

Short answer

The energy of a spherical conductor, of radius a,carrying a charge Qand surrounded by linear dielectric material of susceptibility $Xe$out to radius bis $\frac{Q^2}{8{\mathrm{\pi \epsilon }}_0\left(1+\mathrm{Xe}\right)}\left(\frac{1}{a}+\frac{Xe}{b}\right)$ .

Step-by-step solution

Given data

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.

Electrostatic energy and volume element

The electrostatic energy in a configuration is given by

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\overrightarrow{\mathbf{D}}\mathbf{.}\overrightarrow{\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{\int }\overrightarrow{\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, $\overrightarrow{D}$is the displacement current,$\overrightarrow{E}$is the electric field and$dr$is the infinitesimal volume element.

The infinitesimal volume element for radial symmetry is

$\mathbf{d\tau }\mathbf{=}\mathbf{4}{\mathbf{\tau \tau r}}^{\mathbf{2}}\mathbf{dr}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, $r$is the radial coordinate.

Derivation of energy

The displacement current of the configuration is

$\overrightarrow{D}=\left\{\begin{array}{ll}0& r<a\\ \frac{Q}{4{\mathrm{\pi r}}^2}& r>a\end{array}\right.$

The electric field of the configuration is

$\overrightarrow{E}\left\{\begin{array}{ll}\begin{array}{cc}0& r<a\\ \frac{Q}{4{\mathrm{\pi \epsilon r}}^2}& a<r<b\\ \frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}& r>b\end{array}& \end{array}\right.$

Here, ${\epsilon }_0$is the permittivity of free space and $\epsilon$is the permittivity of the surrounding

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

$$\begin{gathered} E=\frac{1}{2}\frac{Q}{{\left(4\mathrm{\pi }\right)}^2}4\mathrm{\pi }\left\{\frac{1}{3}{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{1}{{\mathrm{r}}^2}\mathrm{dr}+\frac{1}{{\mathrm{\epsilon }}_0}{\int }_{\mathrm{b}}^{\infty }\frac{1}{2}\mathrm{dr}\right\} \\ =\frac{\mathrm{Q}}{8{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{1+\mathrm{Xe}}{\left[-\frac{1}{\mathrm{r}}\right]}_{\mathrm{a}}^{\mathrm{b}}+{\left[-\frac{1}{\mathrm{r}}\right]}_{\mathrm{b}}^{\infty }\right\} \\ =\frac{{\mathrm{Q}}^2}{8{\mathrm{\pi \epsilon }}_0\left(1+\mathrm{Xe}\right)}\left(\frac{1}{\mathrm{a}}+\frac{\mathrm{Xe}}{\mathrm{b}}\right) \end{gathered}$$

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