Chapter 3, Problem 3.30P

(a) Use Maxwell’s first equation. đ�›�. D=P_v, to describe the variation of the electric field intensity with x in a region in which no charge density exists and in which a nonhomogeneous dielectric has a permittivity that increases exponentially with x, The field has an x component on1y; (b) repeat part a, but with a radially directed electric field (spherical coordinates) in which again p_v = 0, but in which the permittivity decrease exponentially with r.

To determine

(a)

Electric field intensity for the given conditions.

Answer to Problem 3.30P

Electric field intensity is $E_0{e}^{-{\alpha }_{1}x}$.

Explanation of Solution

Given Information:

There is no charge density in the region. The nonhomogeneous dielectric permittivity increases exponentially with x.

Concept used:

The expression for permittivity $\epsilon \left(x\right)={\epsilon }_1\exp \left({\alpha }_{1}x\right)$

Using Maxwell's first equation, $\nabla \cdot D={\rho }_v$

General solution for first order non-homogeneous function is $y=c{e}^{-p_x}$

Calculation:

The expression for permittivity $\epsilon \left(x\right)={\epsilon }_1\exp \left({\alpha }_{1}x\right)\left[{\epsilon }_1,{\alpha }_1\text{are}\text{constant}\right]$

Using Maxwell's first equation, $\nabla \cdot D={\rho }_v$

   $\begin{array}{c}\nabla \cdot D=\nabla \left[\epsilon \left(x\right)E\left(x\right)\right]\\ =\nabla \left[{\epsilon }_1{e}^{{\alpha }_{1}x}E\left(x\right)\right]\left[\text{substituting}{\epsilon }_1{e}^{{\alpha }_{1}x}\text{for}\epsilon \left(x\right)\right]\\ =\frac{d}{dx}\left[{\epsilon }_1{e}^{{\alpha }_{1}x}E\left(x\right)\right]\end{array}$

Differentiating the expression with respect to x we have,

   $\begin{array}{c}0={\epsilon }_1\left[{\alpha }_1{e}^{{\alpha }_{1}x}{E}_x+e^{{\alpha }_{1}x}\frac{d}{dx}\left(E_x\right)\right]\\ 0={\alpha }_1{e}^{{\alpha }_{1}x}{E}_x+e^{{\alpha }_{1}x}\frac{d}{dx}\left(E_x\right)\\ 0=e^{{\alpha }_{1}x}\left[\frac{d{E}_x}{dx}+{\alpha }_1{E}_x\right]\\ 0=\frac{d{E}_x}{dx}+{\alpha }_1{E}_x\\ \frac{d{E}_x}{dx}=-{\alpha }_1{E}_x\end{array}$

General solution for first order non-homogeneous function is $y=c{e}^{-p_x}$

Using the expression to find the expression for electric field intensity, $E_x\left(x\right)=E_0{e}^{-{\alpha }_{1}x}$

Here $E_0$ is constant.

Thus, the electric field intensity is $E_0{e}^{-{\alpha }_{1}x}$

To determine

(b)

Electric field intensity for the given conditions.

Answer to Problem 3.30P

   $\frac{E_0}{r^2}{e}^{{\alpha }_{2}r}$

Explanation of Solution

Given Information:

   ${\rho }_v=0$

Permittivity decreases exponentially with r

General solution for first order non-homogeneous function is $y=c.e^{-\rho pdy}$

Concept used:

The expression for permittivity in spherical coordinates is,

   $\epsilon \left(r\right)={\epsilon }_2\exp \left(-{\alpha }_{2}r\right)\left[{\epsilon }_2,{\alpha }_2\text{are}\text{constant}\right]$

Expression for $\nabla \cdot D$ in spherical coordinates is, $\nabla \cdot D=\frac{1}{r^2}\frac{d}{dr}\left[r^2\epsilon \left(r\right)E\left(r\right)\right]$

Calculation:

Expression for $\nabla \cdot D$ in spherical coordinates is, $\nabla \cdot D=\frac{1}{r^2}\frac{d}{dr}\left[r^2\epsilon \left(r\right)E\left(r\right)\right]$

   $\begin{array}{c}\nabla \cdot D=\frac{1}{r^2}\frac{d}{dr}\left[r^2\epsilon \left(r\right)E\left(r\right)\right]\\ =\frac{1}{r^2}\frac{d}{dr}\left[r^2{\epsilon }_2{e}^{-{\alpha }_{2}r}{E}_r\right]\left[\text{substituting}{\epsilon }_2{e}^{-{\alpha }_{2}r}\text{for}\epsilon \left(r\right)\right]\\ =\frac{{\epsilon }_2}{r^2}\frac{d}{dr}\left[r^2{e}^{-{\alpha }_{2}r}{E}_r\right]\\ =\frac{{\epsilon }_2}{r^2}\left[\left(2r{e}^{-{\alpha }_{2}r}{E}_r\right)+\left(-{\alpha }_2{e}^{-{\alpha }_{2}r}\times r^2{E}_r\right)+\left(r^2\times e^{-{\alpha }_{2}r}\times \frac{d{E}_r}{dr}\right)\right]\\ =\frac{{\epsilon }_2}{r^2}\left[\left(2r{E}_r\right)-\left({\alpha }_2\times r^2{E}_r\right)+\left(r^2\times \frac{d{E}_r}{dr}\right)\right]e^{-{\alpha }_{2}r}\end{array}$

Simplifying the expression to find the expression for electric field intensity,

   $\begin{array}{c}0={\epsilon }_2\left[\frac{\left(2r{E}_r\right)}{r^2}-\frac{\left({\alpha }_2{r}^2{E}_r\right)}{r^2}+\left(\frac{d{E}_r}{dr}\right)\right]e^{-{\alpha }_{2}r}\\ 0={\epsilon }_2\left[\frac{\left(2r{E}_r\right)}{r^2}-\frac{\left({\alpha }_2{r}^2{E}_r\right)}{r^2}+\left(\frac{d{E}_r}{dr}\right)\right]\\ 0=\left[\left(\frac{2}{r}-{\alpha }_2\right)E_r+\left(\frac{d{E}_r}{dr}\right)\right]\\ \frac{d{E}_r}{dr}=-\left(\frac{2}{r}-{\alpha }_2\right)E_r\end{array}$

General solution for first order non-homogeneous function is $y=c\cdot e^{-\rho pdy}$

Using this expression to find the expression for electric field intensity

   $\begin{array}{c}{E}_r\left(r\right)=E_0{e}^{-\rho \left(\frac{2}{r}-{\alpha }_2\right)dr}\\ =E_0{e}^{-2\ln \left(r\right)+{\alpha }_{2}r}\\ =E_0\left(e^{-2\ln \left(r\right)}\times e^{{\alpha }_{2}r}\right)\\ =E_0\left(\frac{1}{r^2}\times e^{{\alpha }_{2}r}\right)\\ =\frac{E_0}{r^2}{e}^{{\alpha }_{2}r}\left[\text{Here, }{E}_0\text{is constant}\right]\end{array}$

Conclusion:

Thus, electric field intensity in spherical coordinates is $\frac{E_0}{r^2}{e}^{{\alpha }_{2}r}$