Chapter 6, Problem 69P

(a)

To determine

Calculate the electric field at $\left(2,-2,3\right)$.

Answer to Problem 69P

The electric field at $\left(2,-2,3\right)$ is $\underset{\_}{-138.2a_x-184.3a_y\frac{\text{V}}{\text{m}}}$.

Explanation of Solution

Calculation:

Write the expression for the electric field.

$\begin{array}{c}E=E_{+}+E_{-}\\ =\frac{{\rho }_L}{2\pi {\epsilon }_o}\left(\frac{a_{\rho 1}}{{\left|a_{\rho 1}\right|}^2}-\frac{a_{\rho 2}}{{\left|a_{\rho 2}\right|}^2}\right)\left\{\begin{array}{l}\because {\rho }_1={\left|a_{\rho 1}\right|}^2\\ {\rho }_2={\left|a_{\rho 2}\right|}^2\end{array}\right\}\end{array}$

Substitute 16 nC/m for ${\rho }_L$, $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, $\left(2,-2,3\right)-\left(3,-2,4\right)$ for $a_{\rho 1}$, and $\left(2,-2,3\right)-\left(3,-2,-4\right)$ for $a_{\rho 2}$.

$\begin{array}{c}E=\frac{16\text{nC/m}}{2\pi \left(\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}\right)}\left(\frac{\left(2,-2,3\right)-\left(3,-2,4\right)}{{\left|\left(2,-2,3\right)-\left(3,-2,4\right)\right|}^2}-\frac{\left(2,-2,3\right)-\left(3,-2,-4\right)}{{\left|\left(2,-2,3\right)-\left(3,-2,-4\right)\right|}^2}\right)\frac{1}{\text{m}}\\ =\left(18\times 16\right)\left[\frac{\left(-1,0,-1\right)}{2}-\frac{\left(-1,0,7\right)}{50}\right]\left\{\because \text{C}=\text{F}\cdot \text{V}\right\}\\ =-138.2a_x-184.3a_y\frac{\text{V}}{\text{m}}\end{array}$

Conclusion:

Thus, the electric field at $\left(2,-2,3\right)$ is $\underset{\_}{-138.2a_x-184.3a_y\frac{\text{V}}{\text{m}}}$.

(b)

To determine

Find the induced surface charge density on the conducting plane at $\left(5,-6,0\right)$.

Answer to Problem 69P

The induced surface charge density is $\underset{\_}{-1.018\frac{\text{nC}}{{\text{m}}^2}}$.

Explanation of Solution

Calculation:

Consider the expression for the induced surface charge density.

${\rho }_s=D_n$        (1)

Calculate the displacement field vector.

$\begin{array}{c}D=D_{+}+D_{-}\\ =\frac{{\rho }_L}{2\pi }\left(\frac{a_{\rho 1}}{{\left|a_{\rho 1}\right|}^2}-\frac{a_{\rho 2}}{{\left|a_{\rho 2}\right|}^2}\right)\left\{\begin{array}{l}\because {\rho }_1={\left|a_{\rho 1}\right|}^2\\ {\rho }_2={\left|a_{\rho 2}\right|}^2\end{array}\right\}\end{array}$

Substitute 16 nC/m for ${\rho }_L$, $\left(5,-6,0\right)-\left(3,-6,4\right)$ for $a_{\rho 1}$, and $\left(5,-6,0\right)-\left(3,-6,-4\right)$ for $a_{\rho 2}$.

$\begin{array}{c}D=\frac{16\text{nC/m}}{2\pi }\left(\frac{\left(5,-6,0\right)-\left(3,-6,4\right)}{{\left|\left(5,-6,0\right)-\left(3,-6,4\right)\right|}^2}-\frac{\left(5,-6,0\right)-\left(3,-6,-4\right)}{{\left|\left(5,-6,0\right)-\left(3,-6,-4\right)\right|}^2}\right)\frac{1}{\text{m}}\\ =\left(\frac{8}{\pi }\right)\left[\frac{\left(2,0,-4\right)}{20}-\frac{\left(2,0,4\right)}{20}\right]\frac{\text{nC}}{{\text{m}}^2}\\ =-1.018a_z\frac{\text{nC}}{{\text{m}}^2}\end{array}$

The above equation is compared with the general expression $D=D_n{a}_z$, the value of $D_n$ is $-1.018\frac{\text{nC}}{{\text{m}}^2}$.

Substitute $-1.018\frac{\text{nC}}{{\text{m}}^2}$ for $D_n$ in Equation (1).

${\rho }_s=-1.018\frac{\text{nC}}{{\text{m}}^2}$

Conclusion:

Thus, the induced surface charge density is $\underset{\_}{-1.018\frac{\text{nC}}{{\text{m}}^2}}$.