91Ó°ÊÓ

The space between the plates of a parallel-plate capacitor is filled with dielectric material whose dielectric constant varies linearly from 1 at $x=0$ to 2 at $x=d$.

The capacitor is connected to a battery of voltage V.

The electrostatic potential is

$\mathbf{V}\mathbf{=}\mathbf{-}\mathbf{∫}\stackrel{\mathbf{→}}{\mathbf{E}\mathbf{}}\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{l}}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Volume bound charge density

${\mathbf{ÒÏ}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\stackrel{\mathbf{→}}{\mathbf{∇}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{P}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Surface bound charge density

${\mathit{σ}}_{\mathbf{b}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{P}}\mathbf{.}\hat{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

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

The electrostatic field due to a free charge density ${\mathrm{σ}}_f$at a height $x$ is

$\stackrel{→}{E}=\frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0(1+x/d)}\hat{x}......\left(4\right)$

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space.

Substitute this in equation (1) and get

$$\begin{gathered} V=-{∫}_d^0\frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0(1+x/d)}\hat{x}.d\stackrel{→}{x} \\ =\frac{{\mathrm{σ}}_{f}d}{{\mathrm{Î\mu }}_0}\ln {\left(1+\frac{x}{d}\right)}_0^d \\ =\frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0}\ln 2 \\ {\mathrm{σ}}_f=\frac{{\mathrm{Î\mu }}_{0}v}{d\ln 2} \end{gathered}$$

Substitute this back in equation (4) to get

$$\begin{gathered} \stackrel{→}{E}=\frac{{\displaystyle \frac{{\mathrm{Î\mu }}_{0}V}{d\ln 2}}}{{\mathrm{Î\mu }}_0(1+x/d)}\hat{x} \\ =\frac{V}{d\ln 2(1+x/d)}\hat{x} \end{gathered}$$

The expression for the polarization is

$\stackrel{→}{p}={\mathrm{Î\mu }}_0{x}_e\stackrel{→}{E}$

Here, $X_e$is the susceptibility of the medium which for this setup is $x/d$.

Substitute the value in the above equation

$\stackrel{→}{P}={\mathrm{Î\mu }}_0\frac{Vx}{d^2\ln 2(1+x/d)}\hat{x}$

From equation (2), the volume bound charge density is

$$\begin{gathered} {\mathrm{ÒÏ}}_b=-\stackrel{→}{∇}.\left({\mathrm{Î\mu }}_0\frac{Vx}{d^2\ln 2(1+x/d)}\hat{x}\right) \\ =-\frac{{\mathrm{Î\mu }}_{0}V}{d^2\ln 2}\frac{d}{dx}\frac{x}{1+x/d} \\ =-\frac{{\mathrm{Î\mu }}_{0}V}{d^2\ln 2}\left(\frac{x}{1+x/d}-1\frac{x}{{(1+x/d)}^2}\right) \\ =-\frac{{\mathrm{Î\mu }}_{0}V}{d^2\ln 2}\frac{x}{{(1+x/d)}^2} \end{gathered}$$

From equation (3), the surface bound charge density is

${\mathrm{σ}}_b=\left\{\begin{array}{l}0forx=0\\ \frac{{\mathrm{Î\mu }}_{0}V}{2d\ln 2}forx=d\end{array}\right.$

The net bound charge is then

localid="1657780977126" $$\begin{gathered} Q=∫{\mathrm{ÒÏ}}_{b}d\mathrm{Ï„}+∫{\mathrm{σ}}_{b}da \\ ={∫}_0^d-\frac{{\mathrm{Î\mu }}_{0}V}{d^2\ln 2}\frac{1}{{(1+x/d)}^2}Adx+\frac{{\mathrm{Î\mu }}_{0}V}{2d\ln 2}A \\ =-\frac{{\mathrm{Î\mu }}_{0}VA}{d^2\ln 2}d{\left(-\frac{1}{1+x/d}\right)}_0^d+\frac{{\mathrm{Î\mu }}_{0}V}{2d\ln 2}A \\ =-\frac{{\mathrm{Î\mu }}_{0}V}{2d\ln 2}A+\frac{{\mathrm{Î\mu }}_{0}V}{2d\ln 2}A \\ =0 \end{gathered}$$

Here, A is the susceptibility of the medium which for this setup is .

Thus, the net bound charge is 0.