The space between the plates of a parallel-plate capacitor is filled

with dielectric material whose dielectric constant varies linearly from 1 at the

bottom plate $\mathbf{(}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{)}$to 2 at the top plate $\left(x=d\right)$.The capacitor is connectedto a battery of voltage V.Find all the bound charge, and check that the totalis zero.

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

$\mathbf{V}\mathbf{=}\mathbf{-}\mathbf{\int }\overrightarrow{\mathbf{E}\mathbf{}}\mathbf{d}\overrightarrow{\mathbf{l}}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Volume bound charge density

${\mathbf{\rho }}_{\mathbf{b}}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}\mathbf{.}\overrightarrow{\mathbf{P}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Surface bound charge density

${\mathit{\sigma }}_{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{P}}\mathbf{.}\hat{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

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

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

$\overrightarrow{E}=\frac{{\sigma }_f}{{\epsilon }_0(1+x/d)}\hat{x}......\left(4\right)$

Here, ${\epsilon }_0$is the permittivity of free space.

Substitute this in equation (1) and get

$$\begin{gathered} V=-{\int }_d^0\frac{{\sigma }_f}{{\epsilon }_0(1+x/d)}\hat{x}.d\overrightarrow{x} \\ =\frac{{\sigma }_{f}d}{{\epsilon }_0}\ln {\left(1+\frac{x}{d}\right)}_0^d \\ =\frac{{\sigma }_f}{{\epsilon }_0}\ln 2 \\ {\sigma }_f=\frac{{\epsilon }_{0}v}{d\ln 2} \end{gathered}$$

Substitute this back in equation (4) to get

$$\begin{gathered} \overrightarrow{E}=\frac{{\displaystyle \frac{{\epsilon }_{0}V}{d\ln 2}}}{{\epsilon }_0(1+x/d)}\hat{x} \\ =\frac{V}{d\ln 2(1+x/d)}\hat{x} \end{gathered}$$

The expression for the polarization is

$\overrightarrow{p}={\epsilon }_0{x}_e\overrightarrow{E}$

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

Substitute the value in the above equation

$\overrightarrow{P}={\epsilon }_0\frac{Vx}{d^2\ln 2(1+x/d)}\hat{x}$

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

$$\begin{gathered} {\rho }_b=-\overrightarrow{\nabla }.\left({\epsilon }_0\frac{Vx}{d^2\ln 2(1+x/d)}\hat{x}\right) \\ =-\frac{{\epsilon }_{0}V}{d^2\ln 2}\frac{d}{dx}\frac{x}{1+x/d} \\ =-\frac{{\epsilon }_{0}V}{d^2\ln 2}\left(\frac{x}{1+x/d}-1\frac{x}{{(1+x/d)}^2}\right) \\ =-\frac{{\epsilon }_{0}V}{d^2\ln 2}\frac{x}{{(1+x/d)}^2} \end{gathered}$$

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

${\sigma }_b=\left\{\begin{array}{l}0forx=0\\ \frac{{\epsilon }_{0}V}{2d\ln 2}forx=d\end{array}\right.$

The net bound charge is then

localid="1657780977126" $$\begin{gathered} Q=\int {\rho }_{b}d\tau +\int {\sigma }_{b}da \\ ={\int }_0^d-\frac{{\epsilon }_{0}V}{d^2\ln 2}\frac{1}{{(1+x/d)}^2}Adx+\frac{{\epsilon }_{0}V}{2d\ln 2}A \\ =-\frac{{\epsilon }_{0}VA}{d^2\ln 2}d{\left(-\frac{1}{1+x/d}\right)}_0^d+\frac{{\epsilon }_{0}V}{2d\ln 2}A \\ =-\frac{{\epsilon }_{0}V}{2d\ln 2}A+\frac{{\epsilon }_{0}V}{2d\ln 2}A \\ =0 \end{gathered}$$

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

Thus, the net bound charge is 0.