91Ó°ÊÓ

There is a sphere of radius Rwith frozen-in uniform polarization $\stackrel{→}{P}$.

The expression for the energy of the system is

$$\begin{gathered} \mathbf{W}\mathbf{=}\frac{{\mathbf{Î\mu }}_{\mathbf{0}}}{\mathbf{2}}\mathbf{∫}{\mathbf{E}}^{\mathbf{2}}\mathbf{»åÏ„}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{∫}\stackrel{\mathbf{→}}{\mathbf{D}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{»åÏ„}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Here, is the permittivity of free space, $\stackrel{→}{D}$is the displacement current and $\stackrel{→}{E}$is the electric field.

The infinitesimal volume element in spherical polar coordinates is

$\mathbf{»åÏ„}\mathbf{=}{\mathbf{r}}^{\mathbf{2}}\mathbf{²õ¾\pm ²Ôθ»å°ù»åθ»åÏ•}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here, r , $\mathrm{θ}$and$\mathrm{ϕ}$ are spherical polar coordinates.

The electric field of the configuration is

$\stackrel{→}{E}=\left\{\begin{array}{l}-\frac{\stackrel{→}{P}}{3{\mathrm{Î\mu }}_0}\hat{z}\\ \frac{R^{3}P}{3{\mathrm{Î\mu }}_0{r}^3}\left(2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)\end{array}\right.$

Here, $\stackrel{→}{P}$is the polarization vector and z is a Cartesian coordinate.

Using equation (1) and (3), the work done is

$$\begin{gathered} W=\frac{{\mathrm{Î\mu }}_0}{2}{\left(\frac{\stackrel{→}{P}}{3{\mathrm{Î\mu }}_0}\right)}^2\frac{4}{3}\mathrm{Ï€}{R}^3+\frac{{\mathrm{Î\mu }}_0}{2}{\left(\frac{R^{3}P}{3{\mathrm{Î\mu }}_0}\right)}^{2}2\mathrm{Ï€}{∫}_0^{\mathrm{Ï€}}\left(1+3{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}{∫}_R^{∞}\frac{1}{r^4}dr \\ =\frac{2\mathrm{Ï€}}{27}\frac{P^2{R}^3}{{\mathrm{Î\mu }}_0}+\frac{4\mathrm{Ï€}{R}^3{P}^2}{27{\mathrm{Î\mu }}_0} \\ =\frac{2\mathrm{Ï€}{R}^3{P}^2}{9{\mathrm{Î\mu }}_0} \end{gathered}$$

The displacement current of the configuration is

$$\begin{gathered} \stackrel{→}{D}=\left\{\begin{array}{ll}{\mathrm{Î\mu }}_0\stackrel{→}{E}& r>R\\ {\mathrm{Î\mu }}_0\stackrel{→}{E}+\stackrel{→}{P}& r<R\end{array}\right. \\ =\left\{\begin{array}{ll}{\mathrm{Î\mu }}_0\stackrel{→}{E}& r>R\\ 2{\mathrm{Î\mu }}_0\stackrel{→}{E}& r<R\end{array}\right. \end{gathered}$$

The work done using equation (2) is then

$$\begin{gathered} W=\frac{{\mathrm{Î\mu }}_0}{2}∫E^{2}d\mathrm{Ï„}-2\frac{{\mathrm{Î\mu }}_0}{2}∫E^{2}d\mathrm{Ï„} \\ =\frac{4\mathrm{Ï€}{R}^3{P}^2}{27{\mathrm{Î\mu }}_0}-\frac{4\mathrm{Ï€}{R}^3{P}^2}{27{\mathrm{Î\mu }}_0} \\ =0 \end{gathered}$$

Thus, the work done as calculated using equation (1) is $\frac{2\mathrm{Ï€}{R}^3{P}^2}{9{\mathrm{Î\mu }}_0}$and the work done calculated using (2) is 0. In the second case, the formula for work done is for free charge which is not present in the configuration.