The space between the plates of a parallel-plate capacitor is filled with dielectric material whose dielectirc constant varies linearly from 1 at the bottom plate ($x=0 $) to 2 at the top plate ($x=d $).
The capacitor is connected to a battery of voltage $V $.
Find all the bound charge, and check that the total is zero.
To find the bound charges ($\rho _b $ and $\sigma _b $) we need to know what $\mathbf{P} $ is. To find $\mathbf{P} $, we can use the fact that
\[\mathbf{P}=\epsilon _0 \chi _e \mathbf{E}\tag{Eq. 4.30, page 186} \]So we need to find $\mathbf{E} $. To do so, we use $\mathbf{D}=\epsilon \mathbf{E} $ (Eq. 4.32). We don't have the permittivity, but we know about $\epsilon _r $.
$\epsilon _r $, the dielectric constant (also known as the relative permittivity, see page 187), is defined as
\[\epsilon _r=1+\chi _e=\frac{\epsilon }{\epsilon _0 }\tag{4.34, page 187} \]Since $\epsilon _r(0)=1 $ and $\epsilon _r(d)=2 $, it follows that
\begin{align*} \chi _e(0) &= 0 \\ \chi _e(d) &= 1 \end{align*}The linear function that satisfies this conditions is
\[\chi _e(x)=\frac{x}{d} \]Using Eq. 4.34 again, it follows that
\begin{align*} \text{Permittivity} \;\;\;\epsilon &= \epsilon _0 (1+\chi _e) \\ &= \epsilon _0 \left( 1+\frac{x}{d} \right) \end{align*}Using this in Eq. 4.32 for $\mathbf{E} $, we see that
\[\mathbf{E}=\frac{1}{\epsilon }\mathbf{D}=\frac{1}{\epsilon _0 \left( 1+\frac{x}{d} \right)}\mathbf{D} \]It is clear that $\mathbf{D}=\sigma _f\hat{\mathbf{x}} $, so we conclude that
\[\mathbf{E}=\frac{1}{\epsilon _0 \left( 1+\frac{x}{d} \right) }\sigma _f \hat{\mathbf{x}} \]We can relate $\sigma _f $ to $V $ as follows:
\begin{align*} V &= -\int_{d}^{0} \mathbf{E}\cdot d\mathbf{l} \\ &= \frac{\sigma _f}{\epsilon _0 }\int_{0}^{d} \frac{1}{1+\frac{x}{d}}dx \\ &= \frac{\sigma _f}{\epsilon _0 }d \ln (1+(x/d)) \Big|_{0}^{d} \\ &= \frac{\sigma _f}{\epsilon _0 }d\ln (2)- \frac{\sigma _f}{\epsilon _0 }d \ln (1)\\ &= \frac{\sigma _f}{\epsilon _0 }d\ln (2) \end{align*}So
\[\sigma _f=\frac{\epsilon _0 }{d\ln (2)}V \]This implies that
\[\mathbf{E}=\frac{V}{d \ln (2)}\frac{1}{1+\frac{x}{d}} \hat{\mathbf{x}}\]Now we can go back to Eq. 4.30 and see that
\begin{align*} \mathbf{P} &= \epsilon _0 \chi _e\mathbf{E} \\ &= \epsilon _0 \chi _e\frac{V}{d \ln (2)}\frac{1}{1+\frac{x}{d}} \hat{\mathbf{x}} \\ &= \epsilon _0 \frac{x}{d}\frac{V}{d \ln (2)}\frac{1}{1+\frac{x}{d}} \hat{\mathbf{x}}\\ &= \epsilon _0 \frac{V}{d^2 \ln (2)}\frac{x}{1+\frac{x}{d}} \hat{\mathbf{x}} \end{align*}To find the bound charges, we recall:
\begin{align*} \rho _{\text{b} } &= -\nabla \cdot \mathbf{P} \tag{Eq. 4.12, page 173}\\ \sigma _{\text{b} } &= \mathbf{P}\cdot \hat{\mathbf{n}} \tag{Eq. 4.11, page 173} \end{align*}So:
\begin{align*} \rho _{\text{b} } &= -\frac{V\epsilon _0 }{(d+x)^2 \ln (2)} \\ \sigma _{\text{b} } &= \begin{cases} \begin{matrix} 0 & x=0 \\ \frac{\epsilon _0 }{2d\ln 2}V & x=d \\ \end{matrix} \end{cases} \end{align*}Finally, we verify that the total bound charge is zero (where $S $ denotes surface area):
\begin{align*} Q_{\text{b} } &= \int \rho _bd\tau +\int \sigma _bda \\ &= \int_{0}^{d} -\frac{V\epsilon _0 }{(d+x)^2 \ln (2)}Sdx + \int \frac{\epsilon _0 }{2d\ln 2}VdA\\ &= -\frac{V\epsilon _0}{ \ln 2}S\int_{0}^{d} \frac{1}{(d+x)^2} dx + \frac{\epsilon _0 }{2d\ln 2}VS\\ &= -\frac{V\epsilon _0}{ \ln 2}S\cdot \frac{1}{2d} +\frac{\epsilon _0 }{2d\ln 2}VS\\ &= 0 \end{align*}As expected.