Calculate W,using both Eq. 4.55 and Eq. 4.58, for a sphere of radius

Rwith frozen-in uniform polarization $\overrightarrow{\mathrm{P}}$ (Ex. 4.2). Comment on the discrepancy.

Which (if either) is the "true" energy of the system?

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

The expression for the energy of the system is

$$\begin{gathered} \mathbf{W}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\int }{\mathbf{E}}^{\mathbf{2}}\mathbf{d\tau }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\overrightarrow{\mathbf{D}}\mathbf{.}\overrightarrow{\mathbf{E}}\mathbf{d\tau }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

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

The infinitesimal volume element in spherical polar coordinates is

$\mathbf{d\tau }\mathbf{=}{\mathbf{r}}^{\mathbf{2}}\mathbf{sin\theta drd\theta d\varphi }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here, r , $\theta$and$\varphi$ are spherical polar coordinates.

The electric field of the configuration is

$\overrightarrow{E}=\left\{\begin{array}{l}-\frac{\overrightarrow{P}}{3{\epsilon }_0}\hat{z}\\ \frac{R^{3}P}{3{\epsilon }_0{r}^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)\end{array}\right.$

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

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

$$\begin{gathered} W=\frac{{\epsilon }_0}{2}{\left(\frac{\overrightarrow{P}}{3{\epsilon }_0}\right)}^2\frac{4}{3}\pi R^3+\frac{{\epsilon }_0}{2}{\left(\frac{R^{3}P}{3{\epsilon }_0}\right)}^{2}2\pi {\int }_0^{\mathrm{\pi }}\left(1+3{\cos }^2\theta \right)\sin \theta d\theta {\int }_R^{\infty }\frac{1}{r^4}dr \\ =\frac{2\pi }{27}\frac{P^2{R}^3}{{\epsilon }_0}+\frac{4\pi R^3{P}^2}{27{\epsilon }_0} \\ =\frac{2\pi R^3{P}^2}{9{\epsilon }_0} \end{gathered}$$

The displacement current of the configuration is

$$\begin{gathered} \overrightarrow{D}=\left\{\begin{array}{ll}{\epsilon }_0\overrightarrow{E}& r>R\\ {\epsilon }_0\overrightarrow{E}+\overrightarrow{P}& r<R\end{array}\right. \\ =\left\{\begin{array}{ll}{\epsilon }_0\overrightarrow{E}& r>R\\ 2{\epsilon }_0\overrightarrow{E}& r<R\end{array}\right. \end{gathered}$$

The work done using equation (2) is then

$$\begin{gathered} W=\frac{{\epsilon }_0}{2}\int E^{2}d\tau -2\frac{{\epsilon }_0}{2}\int E^{2}d\tau \\ =\frac{4\pi R^3{P}^2}{27{\epsilon }_0}-\frac{4\pi R^3{P}^2}{27{\epsilon }_0} \\ =0 \end{gathered}$$

Thus, the work done as calculated using equation (1) is $\frac{2\pi R^3{P}^2}{9{\epsilon }_0}$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.