91Ó°ÊÓ

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

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

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

$\stackrel{\mathbf{→}}{\mathbf{D}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\stackrel{\mathbf{→}}{\mathbf{E}}$ …… (1)

Here,${\mathit{Î\mu }}_{\mathbf{0}}$ is relative permittivity and $\stackrel{\mathbf{→}}{\mathbf{E}}$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.

$$\begin{gathered} \stackrel{→}{E}=\stackrel{→}{E_0}-\left(-\frac{\stackrel{→}{P}}{3{\mathrm{Î\mu }}_0}\right) \\ ={\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{3{\mathrm{Î\mu }}_0} \end{gathered}$$

Since the cavity is developed of polarized material.

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

Substitute${\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{3{\mathrm{Î\mu }}_0}$ for $\stackrel{→}{E}$into equation (1).

$$\begin{gathered} \stackrel{→}{D}={\mathrm{Î\mu }}_0{\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{3} \\ =\left({\stackrel{→}{D}}_a=\stackrel{→}{p}\right)+\frac{\stackrel{→}{p}}{3} \\ ={\stackrel{→}{D}}_0-\frac{2}{3}\stackrel{→}{P} \end{gathered}$$

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

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:

$\stackrel{→}{E}≈{\stackrel{→}{E}}_0$

Determine thelong needle-shaped cavity running parallel to P.

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

$$\begin{gathered} \stackrel{→}{D}={\mathrm{Î\mu }}_0{\stackrel{→}{E}}_0 \\ ={\stackrel{→}{D}}_0-\stackrel{→}{P} \end{gathered}$$

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

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

$$\begin{gathered} \stackrel{→}{E}={\stackrel{→}{E}}_0-\frac{\mathrm{σ}b}{{\mathrm{Î\mu }}_0}\stackrel{Áåœ}{p} \\ ={\stackrel{→}{E}}_0-\frac{\left(-P\right)}{{\mathrm{Î\mu }}_0} \\ ={\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{{\mathrm{Î\mu }}_0} \end{gathered}$$

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

Substitute ${\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{{\mathrm{Î\mu }}_0}$for $\stackrel{→}{E}$into equation (1).

$$\begin{gathered} \stackrel{→}{D}={\mathrm{Î\mu }}_0\left({\stackrel{→}{E}}_0+\frac{\stackrel{→}{P}}{{\mathrm{Î\mu }}_0}\right) \\ ={\stackrel{→}{D}}_0 \end{gathered}$$

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