Chapter 6, Problem 59P

(a)

To determine

Calculate the voltage V and electric field E.

Answer to Problem 59P

The voltage V and electric field E are $\underset{\_}{\frac{V_o}{\ln 2}\ln \frac{x+d}{d}\text{and}-\frac{V_o}{\left(x+d\right)\ln 2}{a}_x}$ respectively.

Explanation of Solution

Calculation:

Consider the following expressions.

$V\left(x=0\right)=0\text{V}$

$V\left(x=d\right)=V_o\text{V}$

And

$\epsilon ={\epsilon }_o\left(1+\frac{x}{d}\right)$

From the given data, we know that potential depends only on x. Therefore, Laplace’s equation in Cartesian coordinates is given by,

$\begin{array}{l}\nabla \cdot \left(\epsilon \nabla V\right)=0\\ \frac{d}{dx}\left(\epsilon \nabla V\right)=0\end{array}$

Integrate the above equation with respect to x.

$\begin{array}{l}\epsilon \frac{dV}{dx}=A\\ \frac{dV}{dx}=\frac{A}{\epsilon }\\ \frac{dV}{dx}=\frac{Ad}{{\epsilon }_o\left(x+d\right)}\\ \frac{dV}{dx}=\frac{c_1}{\left(x+d\right)}\left\{\because c_1=\frac{Ad}{{\epsilon }_o}\right\}\end{array}$

Integrate the above equation with respect to x.

$V=c_1\ln \left(x+d\right)+c_2$        (1)

Consider that $V=0$ at $x=0$.

Substitute 0 for x and 0 for V in Equation (1).

$0=c_1\ln \left(0+d\right)+c_2$

$c_2=-c_1\ln \left(d\right)$        (2)

Consider that $V=V_o$ at $x=d$.

Substitute d for x and $V_o$ for V in Equation (1).

$\begin{array}{l}{V}_o=c_1\ln \left(d+d\right)+c_2\\ c_1=\frac{V_o}{\ln 2}\end{array}$

Substitute $\frac{V_o}{\ln 2}$ for $c_1$ in Equation (2).

$c_2=-\frac{V_o}{\ln 2}\ln \left(d\right)$

Substitute $\frac{V_o}{\ln 2}$ for $c_1$ and $-\frac{V_o}{\ln 2}\ln \left(d\right)$ for $c_2$ in Equation (1).

$\begin{array}{c}V=\frac{V_o}{\ln 2}\ln \left(x+d\right)-\frac{V_o}{\ln 2}\ln \left(d\right)\\ =\frac{V_o}{\ln 2}\ln \frac{\left(x+d\right)}{d}\end{array}$

Calculate the electric field.

$\begin{array}{c}E=-\nabla V\\ =-\frac{dV}{dx}{a}_x\\ =-\frac{d\left(\frac{V_o}{\ln 2}\ln \frac{\left(x+d\right)}{d}\right)}{dx}{a}_x\\ =-\frac{V_o}{\left(x+d\right)\ln 2}{a}_x\end{array}$

Conclusion:

Thus, the voltage V and electric field E are $\underset{\_}{\frac{V_o}{\ln 2}\ln \frac{x+d}{d}\text{and}-\frac{V_o}{\left(x+d\right)\ln 2}{a}_x}$ respectively.

(b)

To determine

Find the expression for vector P.

Answer to Problem 59P

The expression for polarization vector P is $\underset{\_}{-\frac{{\epsilon }_{o}x{V}_o}{d\left(x+d\right)\ln 2}{a}_x}$.

Explanation of Solution

Calculation:

Consider the expression for the polarization.

$P={\chi }_e{\epsilon }_{o}E$        (3)

Here,

${\chi }_e$ is electrical susceptibility.

Write the expression for electrical susceptibility ${\chi }_e$.

${\chi }_e={\epsilon }_r-1$

Substitute ${\epsilon }_r-1$ for ${\chi }_e$ in Equation (3).

$P=\left({\epsilon }_r-1\right){\epsilon }_{o}E$

Substitute $\frac{d+x}{d}$ for ${\epsilon }_r$ and $-\frac{V_o}{\left(x+d\right)\ln 2}{a}_x$ for $E$.

$\begin{array}{c}P=\left(\frac{d+x}{d}-1\right){\epsilon }_o\left(-\frac{V_o}{\left(x+d\right)\ln 2}{a}_x\right)\\ =-\frac{{\epsilon }_{o}x{V}_o}{d\left(x+d\right)\ln 2}{a}_x\end{array}$

Conclusion:

Thus, the expression for polarization vector P is $\underset{\_}{-\frac{{\epsilon }_{o}x{V}_o}{d\left(x+d\right)\ln 2}{a}_x}$.

(c)

To determine

Find the density ${\rho }_{Ps}$ at $x=0\text{and}x=d$.

Answer to Problem 59P

The density ${\rho }_{Ps}$ at $x=0\text{and}x=d$ are $\underset{\_}{0\text{and}-\frac{{\epsilon }_o{V}_o}{2d\ln 2}}$ respectively.

Explanation of Solution

Calculation:

Calculate the charge density ${\rho }_{Ps}$ at $x=0$.

$\begin{array}{c}{{\rho }_{Ps}|}_{x=0}={P\cdot \left(a_x\right)|}_{x=0}\\ =\left(-\frac{{\epsilon }_o\left(0\right)V_o}{d\left(x+d\right)\ln 2}{a}_x\right)\left(a_x\right)\\ =0\end{array}$

Calculate the charge density ${\rho }_{Ps}$ at $x=d$.

$\begin{array}{c}{{\rho }_{Ps}|}_{x=d}={P\cdot \left(a_x\right)|}_{x=d}\\ =\left(-\frac{{\epsilon }_o\left(d\right)V_o}{d\left(d+d\right)\ln 2}{a}_x\right)\left(a_x\right)\\ =-\frac{{\epsilon }_o{V}_o}{2d\ln \left(2\right)}\end{array}$

Conclusion:

Thus, the density ${\rho }_{Ps}$ at $x=0\text{and}x=d$ are $\underset{\_}{0\text{and}-\frac{{\epsilon }_o{V}_o}{2d\ln 2}}$ respectively.

(d)

To determine

Find the expression for the capacitance.

Answer to Problem 59P

The expression for capacitance is $\underset{\_}{\frac{{\epsilon }_{o}S}{d\ln 2}}$.

Explanation of Solution

Calculation:

Calculate the electric field.

$\begin{array}{c}E=\frac{{\rho }_s}{\epsilon }{a}_x\\ =\frac{Q}{\epsilon S}{a}_x\left\{\because {\rho }_s=\frac{Q}{S}\right\}\\ =\frac{Q}{{\epsilon }_o\left(1+\frac{x}{d}\right)S}{a}_x\end{array}$

Calculate the voltage of the capacitor.

$\begin{array}{c}V=-{\displaystyle \int E\cdot dl}\\ =-{\displaystyle \underset{a}{\overset{d}{\int }}\frac{Q}{{\epsilon }_{o}S}}\frac{dx}{\left(1+\frac{x}{d}\right)}\\ =\frac{Q}{{\epsilon }_{o}S}d\ln 2\end{array}$

Consider the expression for the capacitance.

$C=\frac{Q}{V}$

Substitute $\frac{Q}{{\epsilon }_{o}S}d\ln 2$ for V.

$\begin{array}{c}C=\frac{Q}{\frac{Q}{{\epsilon }_{o}S}d\ln 2}\\ =\frac{{\epsilon }_{o}S}{d\ln 2}\end{array}$

Conclusion:

Thus, the expression for capacitance is $\underset{\_}{\frac{{\epsilon }_{o}S}{d\ln 2}}$.