Chapter 6, Problem 6.34P

To determine

(a)

The voltage $V$.

Answer to Problem 6.34P

The final voltages are:

$V_1=\frac{-{\rho }_0{z}^2}{2\epsilon }+\frac{{\rho }_{0}bz}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$ and $V_2=\frac{{\rho }_{0}b}{{\epsilon }_0}\left(d-z\right)-\frac{{\rho }_{0}b}{2{\epsilon }_0}\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\left(d-z\right)$.

Explanation of Solution

Calculation:

The Poisson's equation is defined for $z<b$ boundary condition (where $b$ is the radius of conductor).

The Poisson's equation is written as,

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

Here,

${\rho }_0$ is the charge density.

$\epsilon$ is the permittivity.

$z$ is the direction along $z$ axis.

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

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

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

The Laplace's equation is written as,

$\frac{d^2{V}_2}{d{z}^2}=0$    ...... (3)

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

$\frac{d{V}_2}{dz}=C_1{}^{\prime }$    ...... (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,

$\epsilon \frac{d{V}_1}{dz}=-{\rho }_{0}b+\epsilon C_1$    ...... (6)

Equation (3) is multiplied with ${\epsilon }_0$ on both side as,

${\epsilon }_0\frac{d{V}_2}{dz}={\epsilon }_0{C}_1{}^{\prime }$    ...... (7)

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

$-{\rho }_{0}b+\epsilon C_1={\epsilon }_0{C}_1{}^{\prime }$ ..... (8)

Simplified the equation (8) as,

$C_1{}^{\prime }=\frac{-{\rho }_{0}b}{{\epsilon }_0}+\frac{\epsilon C_1}{{\epsilon }_0}$

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

$\frac{d{V}_2}{dz}=\frac{-{\rho }_{0}b}{{\epsilon }_0}+\frac{\epsilon C_1}{{\epsilon }_0}$ ..... (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,

$V_2=\frac{-{\rho }_{0}bz}{{\epsilon }_0}+\frac{\epsilon C_{1}z}{{\epsilon }_0}+C_2{}^{\prime }$ ..... (11)

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

$0=\frac{-{\rho }_{0}bd}{{\epsilon }_0}+\frac{\epsilon C_{1}d}{{\epsilon }_0}+C_2{}^{\prime }$ ..... (12)

Equation (12) is simplified as,

$C_2{}^{\prime }=\frac{{\rho }_{0}bd}{{\epsilon }_0}-\frac{\epsilon C_{1}d}{{\epsilon }_0}$

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

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

Substitute $\frac{{\rho }_{0}bd}{{\epsilon }_0}-\frac{\epsilon C_{1}d}{{\epsilon }_0}$ for $C_2{}^{\prime }$ in equation (11).

$\begin{array}{c}{V}_2=\frac{-{\rho }_{0}bz}{{\epsilon }_0}+\frac{\epsilon C_{1}z}{{\epsilon }_0}+\frac{{\rho }_{0}bd}{{\epsilon }_0}-\frac{\epsilon C_{1}d}{{\epsilon }_0}\\ =\frac{{\rho }_{0}b}{{\epsilon }_0}\left(d-z\right)-\frac{\epsilon C_1}{{\epsilon }_0}\left(d-z\right)\end{array}$    ...... (14)

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

$\frac{-{\rho }_0{z}^2}{2\epsilon }+C_{1}z=\frac{{\rho }_{0}b}{{\epsilon }_0}\left(d-z\right)-\frac{\epsilon C_1}{{\epsilon }_0}\left(d-z\right)$ ..... (15)

Substitute $b$ for $z$ and $\frac{\epsilon }{{\epsilon }_{\text{r}}}$ for ${\epsilon }_0$ in equation (15).

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

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

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

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

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

Conclusion:

Therefore, the final voltages are $V_1=\frac{-{\rho }_0{z}^2}{2\epsilon }+\frac{{\rho }_{0}bz}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$ and $V_2=\frac{{\rho }_{0}b}{{\epsilon }_0}\left(d-z\right)-\frac{{\rho }_{0}b}{2{\epsilon }_0}\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\left(d-z\right)$.

To determine

(b)

The electric field intensity.

Answer to Problem 6.34P

The electric field ${\text{E}}_1$ is $\left[\frac{{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\right]{\text{a}}_z$ and electric field ${\text{E}}_2$ is $\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]{\text{a}}_z$.

Explanation of Solution

Calculation:

Differentiate the equation (16) with respect to $z$ as,

$\frac{d{V}_1}{dz}=\frac{-{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$

Differentiate the equation (17) with respect to $z$ as,

$\frac{d{V}_2}{dz}=-\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$

The electric field intensity ${\text{E}}_1$ is written as,

${\text{E}}_1=-\frac{d{V}_1}{dz}{\text{a}}_z$    ...... (18)

Here,

${\text{a}}_z$ is the plane along $z$ axis.

Substitute $\frac{-{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$ for $\frac{d{V}_1}{dz}$ in equation (18).

$\begin{array}{c}{\text{E}}_1=-\left[\frac{-{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\right]{\text{a}}_z\\ =\left[\frac{{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\right]{\text{a}}_z\end{array}$

The electric field intensity $E_2$ is written as,

${\text{E}}_2=-\frac{d{V}_2}{dz}{\text{a}}_z$    ...... (19)

Substitute $-\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]$ for $\frac{d{V}_2}{dz}$ in equation (19).

$\begin{array}{c}{\text{E}}_2=-\left[-\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\right]{\text{a}}_z\\ =\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]{\text{a}}_z\end{array}$

Conclusion:

Therefore, the electric field ${\text{E}}_1$ is $\left[\frac{{\rho }_{0}z}{\epsilon }+\frac{{\rho }_{0}b}{2\epsilon }\left[\frac{b+2{\epsilon }_{\text{r}}\left(d-b\right)}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]\right]{\text{a}}_z$ and electric field ${\text{E}}_2$ is $\frac{{\rho }_0{b}^2}{2{\epsilon }_0}\left[\frac{1}{b+{\epsilon }_{\text{r}}\left(d-b\right)}\right]{\text{a}}_z$.