Suppose the field inside a large piece of dielectric is ${\mathit{E}}_{\mathbf{0}}$, so that the electric displacement is ${\mathbf{D}}_{\mathbf{0}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{E}}_{\mathbf{0}}\mathbf{+}\mathbf{P}$.

(a) Now a small spherical cavity (Fig. 4.19a) is hollowed out of the material. Find the field at the center of the cavity in terms of ${\mathit{E}}_{\mathbf{0}}$and $\mathbf{P}$. Also find the displacement at the center of the cavity in terms of ${\mathit{D}}_{\mathbf{0}}$and P. Assume the polarization is "frozen in," so it doesn't change when the cavity is excavated. (b) Do the same for a long needle-shaped cavity running parallel to P (Fig. 4.19b).

(c) Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. 4.19c). Assume the cavities are small enough that P,${\mathit{E}}_{\mathbf{0}}$, and ${\mathit{D}}_{\mathbf{0}}$are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.]

Consider thefield inside a large piece of dielectric is $E_0$.

Consider the electric displacement is${\mathrm{D}}_0={\mathrm{\epsilon }}_0{\mathrm{E}}_0+\mathrm{P}$ .

Write the formula of displacement at the center of the cavity in terms of $D_0$and P.

$\overrightarrow{\mathbf{D}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\overrightarrow{\mathbf{E}}$ …… (1)

Here,${\mathit{\epsilon }}_{\mathbf{0}}$ is relative permittivity and $\overrightarrow{\mathbf{E}}$is electric field.

Determine the ${\mathrm{E}}_0$plus the field of a sphere equally polarized with polarization -P will make up the electric field within the hollowed-out hole.

$$\begin{gathered} \overrightarrow{E}=\overrightarrow{E_0}-\left(-\frac{\overrightarrow{P}}{3{\epsilon }_0}\right) \\ ={\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{3{\epsilon }_0} \end{gathered}$$

Since the cavity is developed of polarized material.

Determine thecenter of the cavity in terms of $E_0$and P.

Substitute${\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{3{\epsilon }_0}$ for $\overrightarrow{E}$into equation (1).

$$\begin{gathered} \overrightarrow{D}={\epsilon }_0{\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{3} \\ =\left({\overrightarrow{D}}_a=\overrightarrow{p}\right)+\frac{\overrightarrow{p}}{3} \\ ={\overrightarrow{D}}_0-\frac{2}{3}\overrightarrow{P} \end{gathered}$$

Therefore, the value of displacement at the center of the cavity in terms of $D_0$and P is$\overrightarrow{D}={\overrightarrow{D}}_0-\frac{2}{3}\overrightarrow{P}$

The thin needle may be modeled as a very thin cylinder, acting as a dipole with a dipole moment antiparallel to the P. The ends are far distant if we measure the field at the center of the thin needle, so:

$\overrightarrow{E}\approx {\overrightarrow{E}}_0$

Determine thelong needle-shaped cavity running parallel to P.

Substitute${\overrightarrow{E}}_0$for$\overrightarrow{E}$into equation (1).

$$\begin{gathered} \overrightarrow{D}={\epsilon }_0{\overrightarrow{E}}_0 \\ ={\overrightarrow{D}}_0-\overrightarrow{P} \end{gathered}$$

Therefore, the value of same for a long needle-shaped cavity running parallel to P is$\overrightarrow{D}={\overrightarrow{D}}_0-\overrightarrow{P}$ .

Determine the field due to the short, fat cylinder is approximately the one of a parallel-plate capacitor:

$$\begin{gathered} \overrightarrow{E}={\overrightarrow{E}}_0-\frac{\sigma b}{{\epsilon }_0}\stackrel{⏜}{p} \\ ={\overrightarrow{E}}_0-\frac{\left(-P\right)}{{\epsilon }_0} \\ ={\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{{\epsilon }_0} \end{gathered}$$

Now determine the same for a thin wafer-shaped cavity perpendicular to P.

Substitute ${\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{{\epsilon }_0}$for $\overrightarrow{E}$into equation (1).

$$\begin{gathered} \overrightarrow{D}={\epsilon }_0\left({\overrightarrow{E}}_0+\frac{\overrightarrow{P}}{{\epsilon }_0}\right) \\ ={\overrightarrow{D}}_0 \end{gathered}$$

Therefore, the value of same for a thin wafer-shaped cavity perpendicular to P is $\overrightarrow{D}={\overrightarrow{D}}_0$.