Question

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?

Short answer

For a sphere of radius Rwith frozen-in uniform polarization $\overrightarrow{\mathit{P}}$using Eq. 4.55 is $\frac{2\pi R^3{P}^2}{9{\epsilon }_0}$and that using Eq. 4.58 is 0. Eq. 4.58 is the work done in the presence of free charge which is absent in the configuration causing the discrepancy in the result.

Step-by-step solution

Given data

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

Energy of the system and volume element

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.

Derivation of work done

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.