Chapter 6, Problem 15P

To determine

Calculate the values of V, E, and D and find the values of density ${\rho }_s$ for $\rho$ values of 5 mm and 15 mm.

Answer to Problem 15P

The values of V, E, and D are $\underset{\_}{\text{36}\text{.91}\text{V}}$, $\underset{\_}{9.102a_{\rho }\text{kV/m}}$, and $\underset{\_}{161a_{\rho }{\text{nC/m}}^2}$ respectively. The values of density ${\rho }_s$ for $\rho$ values of 5 mm and 15 mm are $\underset{\_}{322{\text{nC/m}}^2}$ and $\underset{\_}{-107.3{\text{nC/m}}^2}$ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

$V=\frac{V_o\ln \left(\frac{\rho }{a}\right)}{\ln \left(\frac{b}{a}\right)}$

$V=\frac{V_o\ln \left(\frac{a}{\rho }\right)}{\ln \left(\frac{a}{b}\right)}$        (1)

Write the expression for the electric field.

$E=-\nabla V$

Substitute $\frac{V_o\ln \left(\frac{a}{\rho }\right)}{\ln \left(\frac{a}{b}\right)}$ for V.

$\begin{array}{c}E=-\nabla \left(\frac{V_o\ln \left(\frac{a}{\rho }\right)}{\ln \left(\frac{a}{b}\right)}\right)\\ =-\frac{V_o}{\rho \ln \left(\frac{b}{a}\right)}{a}_{\rho }\end{array}$

$E=\frac{V_o}{\rho \ln \left(\frac{a}{b}\right)}{a}_{\rho }$        (2)

Write the expression for the displacement field vector.

$D=\epsilon E$

Substitute ${\epsilon }_o{\epsilon }_r$ for $\epsilon$ and $-\frac{V_o}{\rho \ln \left(\frac{b}{a}\right)}{a}_{\rho }$ for E.

$D={\epsilon }_o{\epsilon }_r\left(-\frac{V_o}{\rho \ln \left(\frac{b}{a}\right)}{a}_{\rho }\right)$

$D=-{\epsilon }_o{\epsilon }_r\frac{V_o}{\rho \ln \left(\frac{b}{a}\right)}{a}_{\rho }$        (3)

Write the expression for the expression for density.

${\rho }_s=\pm {\epsilon }_o{\epsilon }_r\frac{V_o}{\rho \ln \left(\frac{a}{b}\right)}$        (4)

Consider that $V_o=100\text{V}$, $b=5\text{mm}$, $a=15\text{mm}$, and ${\epsilon }_r=2$.

Substitute 100 V for $V_o$, 15 mm for a, 10 mm for $\rho$, and 5 mm for b in Equation (1).

$\begin{array}{c}V=\frac{100\ln \left(\frac{15}{10}\right)}{\ln \left(\frac{15}{5}\right)}\\ =36.91\text{V}\end{array}$

Substitute 100 V for $V_o$, 15 mm for a, 10 mm for $\rho$, and 5 mm for b in Equation (2).

$\begin{array}{c}E=\frac{100\text{V}}{\left(10\text{mm}\right)\ln \left(\frac{15\text{mm}}{5\text{mm}}\right)}{a}_{\rho }\\ =9.102a_{\rho }\text{kV/m}\end{array}$

Substitute 2 for ${\epsilon }_r$, $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, 100 V for $V_o$, 15 mm for a, 10 mm for $\rho$, and 5 mm for b in Equation (3).

$\begin{array}{c}D=-\left(\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}\right)\left(2\right)\frac{100\text{V}}{\left(10\text{mm}\right)\ln \left(\frac{5\text{mm}}{15\text{mm}}\right)}{a}_{\rho }\\ =161a_{\rho }\times {10}^{-9}{\text{C/m}}^2\left\{\because 1\text{C}=1\text{F}\cdot 1\text{V}\right\}\\ =161a_{\rho }{\text{nC/m}}^2\end{array}$

Substitute 2 for ${\epsilon }_r$, $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, 100 V for $V_o$, 15 mm for a, 5 mm for $\rho$, and 5 mm for b in Equation (4) to find the density ${\rho }_s$ when $\rho =5\text{mm}$.

$\begin{array}{c}{\rho }_s=\left(\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}\right)\left(2\right)\frac{100\text{V}}{\left(5\text{mm}\right)\ln \left(\frac{15\text{mm}}{5\text{mm}}\right)}\\ =322\times {10}^{-9}{\text{C/m}}^2\\ =322{\text{nC/m}}^2\end{array}$

Substitute 2 for ${\epsilon }_r$, $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, 100 V for $V_o$, 15 mm for a, 5 mm for $\rho$, and 15 mm for b in Equation (4) to find the density ${\rho }_s$ when $\rho =15\text{mm}$.

$\begin{array}{c}{\rho }_s=-\left(\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}\right)\left(2\right)\frac{100\text{V}}{\left(15\text{mm}\right)\ln \left(\frac{15\text{mm}}{5\text{mm}}\right)}\\ =-107.3\times {10}^{-9}{\text{C/m}}^2\\ =-107.3{\text{nC/m}}^2\end{array}$

Conclusion:

Thus, the values of V, E, and D are $\underset{\_}{\text{36}\text{.91}\text{V}}$, $\underset{\_}{9.102a_{\rho }\text{kV/m}}$, and $\underset{\_}{161a_{\rho }{\text{nC/m}}^2}$ respectively. The values of density ${\rho }_s$ for $\rho$ values of 5 mm and 15 mm are $\underset{\_}{322{\text{nC/m}}^2}$ and $\underset{\_}{-107.3{\text{nC/m}}^2}$ respectively.