Chapter 5, Problem 40P

To determine

Calculate the value of $D_2$ and the angle ${\theta }_2$ that $D_2$ makes with the normal.

Answer to Problem 40P

The value of $D_2$ is $5.96a_x-9.28a_y+16.25a_z\text{nC}/{\text{m}}^{\text{2}}$ and the angle ${\theta }_2$ that $D_2$ makes with the normal is $87.66°$.

Explanation of Solution

Calculation:

Given, the electric interface is defined by $4x+3y=10\text{m}$ and the electric flux density is

$D_1=2a_x-4a_y+6.5a_z\text{nC}/{\text{m}}^{\text{2}}$

Find the unit vector normal to the interface as follows,

$\begin{array}{c}{a}_n=\frac{4a_x+3a_y}{\sqrt{4^2+3^2}}\\ =\frac{4a_x+3a_y}{5}\\ =0.8a_x+0.6a_y\end{array}$

Consider the general expression for dielectric-dielectric interface.

$D_{1n}=D_{2n}$        (1)

Find the value $D_{1n}$.

$D_{1n}=D_1\cdot a_n$

Substitute $\left(2a_x-4a_y+6.5a_z\right)\times {10}^{-9}$ for $D_1$ and $0.8a_x+0.6a_y$ for $a_n$ in above expression.

$\begin{array}{c}{D}_{1n}=\left(2a_x-4a_y+6.5a_z\right)\times {10}^{-9}\left[\left(0.8a_x+0.6a_y\right)\right]\\ =-0.8\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Therefore,

$D_{1n}=D_{1n}\cdot a_n$

Substitute $\left(-0.8\right)\times {10}^{-9}$ for $D_{1n}$ and $0.8a_x+0.6a_y$ for $a_n$ in above expression

$\begin{array}{c}{D}_{1n}=\left(-0.8\right)\times {10}^{-9}\left(0.8a_x+0.6a_y\right)\\ =\left(-0.64a_x-0.48a_y\right)\times {10}^{-9}\\ =-0.64a_x-0.48a_y\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Therefore, from Equation (1),

$D_{1n}=D_{2n}=-0.64a_x-0.48a_y\text{nC}/{\text{m}}^{\text{2}}$

Find the value of $D_{1t}$.

$D_{1t}=D_1-D_{1n}$

Substitute $\left(2a_x-4a_y+6.5a_z\right)\times {10}^{-9}$ for $D_1$ and $\left(-0.64a_x-0.48a_y\right)\times {10}^{-9}$ for $D_{1n}$ in above equation.

$\begin{array}{c}{D}_{1t}=\left[\left(2a_x-4a_y+6.5a_z\right)\times {10}^{-9}\right]-\left[\left(-0.64a_x-0.48a_y\right)\times {10}^{-9}\right]\\ =\left(2.64a_x-3.52a_y+6.5a_z\right)\times {10}^{-9}\\ =\left(2.64a_x-3.52a_y+6.5a_z\right)\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Consider the general expression for dielectric-dielectric interface.

$E_{1t}=E_{2t}$

$\frac{D_{1t}}{{\epsilon }_{r1}}=\frac{D_{2t}}{{\epsilon }_{r2}}$        (2)

As region including the origin in free space, ${\epsilon }_{r1}=1$.

Substitute 1 for ${\epsilon }_{r1}$, 2.5 for ${\epsilon }_{r2}$, and $\left(2.64a_x-3.52a_y+6.5a_z\right)\times {10}^{-9}$ for $D_{1t}$ in Equation (2).

$\begin{array}{l}\frac{\left(2.64a_x-3.52a_y+6.5a_z\right)\times {10}^{-9}}{1}=\frac{D_{2t}}{2.5}\\ D_{2t}=2.5\left(2.64a_x-3.52a_y+6.5a_z\right)\times {10}^{-9}\\ D_{2t}=6.6a_x-8.8a_y+16.25a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Find the value of $D_2$.

$D_2=D_{2t}+D_{2n}$

Substitute $\left(6.6a_x-8.8a_y+16.25a_z\right)\times {10}^{-9}$ for $D_{2t}$ and $\left(-0.64a_x-0.48a_y\right)\times {10}^{-9}$ for $D_{2n}$ in above expression.

$\begin{array}{c}{D}_2=\left[\left(6.6a_x-8.8a_y+16.25a_z\right)\times {10}^{-9}\right]+\left[\left(-0.64a_x-0.48a_y\right)\times {10}^{-9}\right]\\ =\left(5.96a_x-9.28a_y+16.25a_z\right)\times {10}^{-9}\\ =5.96a_x-9.28a_y+16.25a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Consider the general expression to find the angle ${\theta }_2$ that $D_2$ makes with the normal.

$\begin{array}{c}\tan {\theta }_2=\frac{\left|D_{2t}\right|}{\left|D_{2n}\right|}\\ =\frac{\sqrt{{6.6}^2+{8.8}^2+{16.25}^2}}{\sqrt{{0.64}^2+{0.48}^2}}\\ =24.53\\ {\theta }_2={\tan }^{-1}\left(24.53\right)\end{array}$

The above equation becomes,

${\theta }_2=87.66°$

Conclusion:

Thus, the value of $D_2$ is $5.96a_x-9.28a_y+16.25a_z\text{nC}/{\text{m}}^{\text{2}}$ and the angle ${\theta }_2$ that $D_2$ makes with the normal is $87.66°$.