Chapter 5, Problem 5.22P

The line segment x=0, -1≤y≤1, z=1, carries a linear charge $p_L=x\left|y\right|\mu C/m,$

Let z=0 be a conducting plane, and determine the surface charge density at; (a0 (0,0,0); (b) (0,1,0).

To determine

(a)

The surface charge density at $\left(0,0,0\right)$.

Answer to Problem 5.22P

The surface charge density at $\left(0,0,0\right)$ is, ${\rho }_S=-0.29\mu {\text{C/m}}^2$.

Explanation of Solution

Given Information:

The line segment $x=0,-1\le y\le 1,z=1$ , is having a linear charge density ${\rho }_L=\pi \left|y\right|\mu \text{C/m}$ . $z=0$ is a conducting plane.

Calculation:

Consider the line charge is made up of different segments. Each segment have a length of $dy$ , and so charge $dq={\rho }_{L}dy$ . For a segment at $\left(0,y,1\right)$ , corresponding image charge segment is at $\left(0,y,-1\right)$ . So, the flux density on the given point that associated with each segment image charge pair,

   $\begin{array}{l}d\text{D}=\frac{{\rho }_{L}dy\left[\left(0-y\right){\text{a}}_y-{\text{a}}_z\right]}{4\pi {\left[{\left(0-y\right)}^2+1\right]}^{3/2}}-\frac{{\rho }_{L}dy\left[\left(0-y\right){\text{a}}_y+{\text{a}}_z\right]}{4\pi {\left[{\left(0-y\right)}^2+1\right]}^{3/2}}\\ =-\frac{{\rho }_{L}dy}{2\pi {\left[y^2+1\right]}^{3/2}}{\text{a}}_z\end{array}$

So, the total flux density,

   $\begin{array}{l}\text{D}={\displaystyle \int d\text{D}}={\displaystyle \underset{-1}{\overset{1}{\int }}-\frac{{\rho }_{L}dy}{2\pi {\left[y^2+1\right]}^{3/2}}{\text{a}}_z}={\displaystyle \underset{-1}{\overset{1}{\int }}-\frac{\pi \left|y\right|dy}{2\pi {\left[y^2+1\right]}^{3/2}}{\text{a}}_z}\\ =-{\text{a}}_z{\displaystyle \underset{0}{\overset{1}{\int }}\frac{y}{{\left[y^2+1\right]}^{3/2}}}dy\because u=y^2+1\Rightarrow ydy=du/2\\ =-\frac{{\text{a}}_z}{2}{\displaystyle \underset{1}{\overset{2}{\int }}{u}^{-3/2}}du=\frac{{\text{a}}_z}{2}{\left[\frac{2}{\sqrt{u}}\right]}_1^2\\ =-0.29{\text{a}}_z\end{array}$

Thus, surface charge density,

   $\begin{array}{l}{\rho }_S=D\cdot {\text{a}}_z=-0.29{\text{a}}_z\cdot {\text{a}}_z\\ =-0.29\mu {\text{C/m}}^2\end{array}$

To determine

(b)

The surface charge density at $\left(0,1,0\right)$.

Answer to Problem 5.22P

The surface charge density at $\left(0,1,0\right)$ is, ${\rho }_S=-0.24\mu {\text{C/m}}^2$.

Explanation of Solution

Given Information:

The line segment $x=0,-1\le y\le 1,z=1$ , is having a linear charge density ${\rho }_L=\pi \left|y\right|\mu \text{C/m}$ . $z=0$ is a conducting plane.

Calculation:

Consider the line charge is made up of different segments. Each segment have a length of $dy$ , and so charge $dq={\rho }_{L}dy$ . For a segment at $\left(0,y,1\right)$ , corresponding image charge segment is at $\left(0,y,-1\right)$ . So, the flux density on the given point that associated with each segment image charge pair,

   $\begin{array}{l}d\text{D}=\frac{{\rho }_{L}dy\left[\left(1-y\right){\text{a}}_y-{\text{a}}_z\right]}{4\pi {\left[{\left(1-y\right)}^2+1\right]}^{3/2}}-\frac{{\rho }_{L}dy\left[\left(1-y\right){\text{a}}_y+{\text{a}}_z\right]}{4\pi {\left[{\left(1-y\right)}^2+1\right]}^{3/2}}\\ =-\frac{{\rho }_{L}dy}{2\pi {\left[{\left(1-y\right)}^2+1\right]}^{3/2}}{\text{a}}_z\end{array}$

So, the total flux density,

   $\begin{array}{l}\text{D}={\displaystyle \int d\text{D}}={\displaystyle \underset{-1}{\overset{1}{\int }}-\frac{{\rho }_{L}dy}{2\pi {\left[{\left(1-y\right)}^2+1\right]}^{3/2}}{\text{a}}_z}={\displaystyle \underset{-1}{\overset{1}{\int }}-\frac{\pi \left|y\right|dy}{2\pi {\left[{\left(1-y\right)}^2+1\right]}^{3/2}}{\text{a}}_z}\\ =-{\text{a}}_z{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{y}{2{\left[{\left(1-y\right)}^2+1\right]}^{3/2}}}dy\\ =-{\text{a}}_z{\left[\frac{\left(y-2\right)y}{2\sqrt{{\left(1-y\right)}^2+1}\left|y\right|}\right]}_{-1}^1\\ =-0.24{\text{a}}_z\end{array}$

Thus, surface charge density,

   $\begin{array}{l}{\rho }_S=D\cdot {\text{a}}_z=-0.41{\text{a}}_z\cdot {\text{a}}_z\\ =-0.24\mu {\text{C/m}}^2\end{array}$