Chapter 6, Problem 6.31P

For the parallel-plate capacitor shown in Figure 6. 3, find the potential field in the interior if the upper plate (at z = d) is raised to potential V0, while the lower plate (at z = 0) is grounded. Do this by solving Laplace' s equation separately in each of the two dielectrics. These solutions, as well as the electric flux density, must be continuous across the dielectric interface. Take the interface to he at z = b.

To determine

The potential field between plates $V$.

Answer to Problem 6.31P

The values are $V_1$ is $\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+V_0$ and $V_2$ is $\frac{{\rho }_{0}d}{{\epsilon }_0}\left(b-z\right)-V_0\frac{\left(b-z\right)}{z}$.

Explanation of Solution

Calculation:

The Poisson's equation is defined for $z<d$ boundary condition (where, $d$ is the distance between two plates).

The Poisson's equation (generalization of Laplace equation) is written as

   $\frac{d^2{V}_1}{d{z}^2}=\frac{-{\rho }_0}{{\epsilon }_1}$ ...... (1)

Here,

   ${\rho }_0$ is the charge density.

   ${\epsilon }_1$ is the relative permittivity.

   $z$ is the direction along $z$ -axis

   $V_1$ is the voltage for case $z<b$.

Integrate the equation (1) with respect to $z$.

   $\frac{d{V}_1}{dz}=\frac{-{\rho }_{0}z}{{\epsilon }_1}+C_1$ ...... (2)

The Poisson's equation is defined for the region $b<z<d$.

The Poisson's equation is written as

   $\frac{d^2{V}_2}{d{z}^2}=\frac{-{\rho }_0}{{\epsilon }_2}$ ...... (3)

Integrate the equation (3) with respect to $z$.

   $\frac{d{V}_2}{dz}=\frac{-{\rho }_0\left(d-b\right)}{{\epsilon }_2}+C_1{}^{\prime }$ ..... (4)

Substitute $z=b$ in equation (2).

   $\frac{d{V}_1}{dz}=\frac{-{\rho }_{0}b}{\epsilon }+C_1$ ...... (5)

Simplified the equation (5) as

   ${\epsilon }_1\frac{d{V}_1}{dz}=-{\rho }_{0}b+{\epsilon }_1{C}_1$ ...... (6)

Equation (4) is multiplied with ${\epsilon }_0$ on both side as

   ${\epsilon }_2\frac{d{V}_2}{dz}=-{\rho }_0\left(d-b\right)+{\epsilon }_2{C}_1{}^{\prime }$ ...... (7)

If equation (6) and equation (7) is equal, then it is written as

   $-{\rho }_{0}b+{\epsilon }_1{C}_1=-{\rho }_0\left(d-b\right)+{\epsilon }_2{C}_1{}^{\prime }$ ..... (8)

Simplified the equation (8) as

   $C_1{}^{\prime }=\frac{-{\rho }_{0}b}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}+\frac{{\rho }_0\left(d-b\right)}{{\epsilon }_2}$

Substitute $\frac{-{\rho }_{0}b}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}+\frac{{\rho }_0\left(d-b\right)}{{\epsilon }_2}$ for $C_1{}^{\prime }$ in equation (4).

   $\begin{array}{c}\frac{d{V}_2}{dz}=\frac{-{\rho }_0\left(d-b\right)}{{\epsilon }_2}+\frac{-{\rho }_{0}b}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}+\frac{{\rho }_0\left(d-b\right)}{{\epsilon }_2}\\ =\frac{-{\rho }_{0}b}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}\end{array}$ ..... (9)

Integrate the equation (2) with respect to $z$ as

   $V_1=\frac{-{\rho }_0{z}^2}{2\epsilon }+C_{1}z+C_2$ ...... (10)

Integrate the equation (9) with respect to $z$ as

   $V_2=\frac{-{\rho }_{0}bz}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_{1}z}{{\epsilon }_2}+C_2{}^{\prime }$ ..... (11)

Substitute $b$ for $z$ and $0$ for $V_2$ in equation (11).

   $0=\frac{-{\rho }_0{b}^2}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_{1}b}{{\epsilon }_2}+C_2{}^{\prime }$ ..... (12)

Equation (12) is simplified as

   $C_2{}^{\prime }=\frac{{\rho }_0{b}^2}{{\epsilon }_2}-\frac{{\epsilon }_1{C}_{1}b}{{\epsilon }_2}$

Substitute $0$ for $C_2$ in equation (10).

   $V_1=\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+C_{1}z$ ...... (13)

Substitute $\frac{{\rho }_0{b}^2}{{\epsilon }_2}-\frac{{\epsilon }_1{C}_{1}b}{{\epsilon }_2}$ for $C_2{}^{\prime }$ in equation (11).

   $\begin{array}{c}{V}_2=\frac{-{\rho }_{0}bz}{{\epsilon }_2}+\frac{{\epsilon }_1{C}_{1}z}{{\epsilon }_2}+\frac{{\rho }_0{b}^2}{{\epsilon }_2}-\frac{{\epsilon }_1{C}_{1}b}{{\epsilon }_2}\\ =\frac{{\rho }_{0}b}{{\epsilon }_2}\left(b-z\right)-\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}\left(b-z\right)\end{array}$ ...... (14)

If equation (13) and equation (14) is equal, then it is written as

   $\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+C_{1}z=\frac{{\rho }_{0}b}{{\epsilon }_2}\left(b-z\right)-\frac{{\epsilon }_1{C}_1}{{\epsilon }_2}\left(b-z\right)$ ..... (15)

Substitute $d$ for $z$ in equation (15).

   $C_1=\frac{{\rho }_{0}d}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$

Substitute $\frac{{\rho }_{0}d}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$ for $C_1$ in equation (13)

   $V_1=\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+\frac{{\rho }_{0}dz}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$ ...... (16)

Substitute $\frac{{\rho }_{0}d}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$ for $C_1$ in equation (14)

   $V_2=\frac{{\rho }_{0}d}{{\epsilon }_0}\left(b-z\right)-\frac{{\rho }_{0}d}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]\left(b-z\right)$ ...... (17)

The potential difference $V_0$ is

   $V_0=\frac{{\rho }_{0}dz}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$

Substitute $V_0$ for $\frac{{\rho }_{0}dz}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$ in equation (16).

   $V_1=\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+V_0$

Substitute $\frac{{\rho }_{0}dz}{2{\epsilon }_2}\left[\frac{d+2{\epsilon }_1\left(b-d\right)}{d+{\epsilon }_1\left(b-d\right)}\right]$ for $V_0$ in equation (17).

   $V_2=\frac{{\rho }_{0}d}{{\epsilon }_0}\left(b-z\right)-V_0\frac{\left(b-z\right)}{z}$

Conclusion:

Therefore, the final voltage $V_1$ is $\frac{-{\rho }_0{z}^2}{2{\epsilon }_2}+V_0$ and $V_2$ is $\frac{{\rho }_{0}d}{{\epsilon }_0}\left(b-z\right)-V_0\frac{\left(b-z\right)}{z}$.