Question

A dielectric cube of side a,centered at the origin, carries a "frozen in"

polarization $\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{k}\overrightarrow{\mathbf{r}}$, where kis a constant. Find all the bound charges, and check

that they add up to zero.

Short answer

The net bound charge in a dielectric cube of side a,centered at the origin, carrying a polarization $\overrightarrow{\mathrm{p}}=\mathrm{k}\overrightarrow{\mathrm{r}}$is 0.

Step-by-step solution

Given data

There is a dielectric cube of side a,centered at the origin.

The cube carries a polarization $\overrightarrow{p}=k\overrightarrow{r}.$

Volume and surface bound charge density

Volume bound charge density

${\mathbf{p}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}\mathbf{.}\overrightarrow{\mathbf{P}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\overrightarrow{p}$ is polarization.

Surface bound charge density

${\mathbf{\sigma }}_{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{P}}\mathbf{.}\hat{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, $\hat{n}$is the unit vector along the normal to the surface.

Net bound charge in the cube

The polarization is

$$\begin{gathered} \overrightarrow{P}=K\overrightarrow{r} \\ =k(x\hat{x}+y\hat{y}+z\hat{z}) \end{gathered}$$

From equation (1), the volume bound charge density is

$$\begin{gathered} {\rho }_b=-\overrightarrow{\nabla }.\left(k(x\hat{x}+y\hat{y}+z\hat{z})\right) \\ =-3k \end{gathered}$$

Total volume charge from the volume charge density is

$$\begin{gathered} Q_{\rho }={\rho }_b{a}^3 \\ =-3k{a}^3 \end{gathered}$$

Let the top surface be perpendicular to the z axis.

Thus,

$\hat{n}=\hat{z}$and $z=a/2$

Thus, from equation (2), the surface charge density on the top surface is

role="math" localid="1657775319786" $$\begin{gathered} {\sigma }_b=k(x\hat{x}+y\hat{y}+z\hat{z}).\hat{z} \\ =kz \\ =\frac{ka}{2} \end{gathered}$$

The total charge on the top surface is

$$\begin{gathered} {\sigma }_b={\sigma }_b{a}^2 \\ =k{a}^3 \end{gathered}$$

The surface charges are equal on all the six surfaces. The total bound charge in the cube is then

$$\begin{gathered} Q_b=Q_{\rho }+6Q_{\sigma } \\ =-3k{a}^3+6\times \frac{k{a}^3}{2} \\ =0 \end{gathered}$$

Thus, the net bound charge in the cube is 0.