Chapter 4, Problem 18P

(a)

To determine

The electric field at $P\left(5,-1,4\right)$.

Explanation of Solution

Given:

Surface charge density $\left({\rho }_x\right)$ at $x=2$ is $10\mu \text{C}/{\text{m}}^{\text{2}}$.

Surface charge density $\left({\rho }_y\right)$ at $y=-3$ is $-20\mu \text{C}/{\text{m}}^{\text{2}}$.

Surface charge density $\left({\rho }_z\right)$ at $z=5$ is $30\mu \text{C}/{\text{m}}^2$.

Calculation:

Calculate the electric field $\left(E_5\right)$ at point $x=5$  using the relation.

  $E_5=\frac{{\rho }_x}{2{\epsilon }_0}{a}_x$

  $\begin{array}{c}{E}_5=\frac{10\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}{a}_x\\ =\left(565486.6\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_x\\ =565.486a_x\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_{-1}\right)$ at point $y=-1$  using the relation.

  $E_{-1}=\frac{{\rho }_y}{2{\epsilon }_0}{a}_y$

  $\begin{array}{c}{E}_{-1}=\frac{-20\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}{a}_y\\ =\left(-1130.97\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_y\\ =-1130.97a_y\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_4\right)$ at point $z=4$ using the relation.

  $E_4=\frac{{\rho }_z}{2{\epsilon }_0}{a}_z$

  $\begin{array}{c}{E}_4=\frac{30\mu \text{C}/{\text{m}}^2\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}\left(-a_z\right)\\ =\left(-1696.46\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_z\\ =-1696.46a_z\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_P\right)$ at point $P\left(5,-1,4\right)$ using the relation.

  $E_P=E_5+E_{-1}+E_4$

  $E_P=565.486a_x\text{kN}/\text{C}-1130.97a_y\text{kN}/\text{C}-1696.46a_z\text{kN}/\text{C}$

Thus, the electric field at $P\left(5,-1,4\right)$ is $\left(565.486a_x-1130.97a_y-1696.46a_z\right)\text{kN}/\text{C}$.

(b)

To determine

The electric field at $R\left(0,-2,1\right)$.

Explanation of Solution

Calculation:

Calculate the electric field $\left(E_0\right)$ at point $x=0$  using the relation.

  $E_0=\frac{{\rho }_x}{2{\epsilon }_0}\left(-a_x\right)$

  $\begin{array}{c}{E}_0=\frac{10\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}\left(-a_x\right)\\ =\left(-565486.6\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_x\\ =-565.486a_x\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_{-2}\right)$ at point $y=-2$  using the relation.

  $E_{-2}=\frac{{\rho }_y}{2{\epsilon }_0}\left(-a_y\right)$

  $\begin{array}{c}{E}_{-2}=\frac{-20\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}\left(-a_y\right)\\ =\left(1130.97\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_y\\ =1130.97a_y\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_1\right)$ at point $z=1$  using the relation.

  $E_1=\frac{{\rho }_z}{2{\epsilon }_0}\left(-a_z\right)$

  $\begin{array}{c}{E}_1=\frac{30\mu \text{C}/{\text{m}}^2\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}\left(-a_z\right)\\ =\left(1696.46\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)\left(-a_z\right)\\ =-1696.46a_z\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_R\right)$ at point $R\left(0,-2,1\right)$ using the relation.

  $E_R=E_5+E_{-1}+E_4$

  $E_R=-565.486a_x\text{kN}/\text{C}-1130.97a_y\text{kN}/\text{C}-1696.46a_z\text{kN}/\text{C}$

Thus, the electric field at $R\left(0,-2,1\right)$ is $\left(-565.486a_x-1130.97a_y-1696.46a_z\right)\text{kN}/\text{C}$.

(c)

To determine

The electric field at $Q\left(3,-4,10\right)$.

Explanation of Solution

Calculation:

Calculate the electric field $\left(E_3\right)$ at point $x=3$  using the relation.

  $E_3=\frac{{\rho }_x}{2{\epsilon }_0}{a}_x$

  $\begin{array}{c}{E}_3=\frac{10\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}{a}_x\\ =\left(565486.6\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_x\\ =565.486a_x\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_{-4}\right)$ at point $y=-4$  using the relation.

  $E_{-4}=\frac{{\rho }_y}{2{\epsilon }_0}\left(-a_y\right)$

  $\begin{array}{c}{E}_{-4}=\frac{-20\mu \text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}\left(-a_y\right)\\ =\left(1130.97\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_y\\ =1130.97a_y\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_{10}\right)$ at point $z=10$ using the relation.

  $E_{10}=\frac{{\rho }_z}{2{\epsilon }_0}{a}_z$

  $\begin{array}{c}{E}_{10}=\frac{30\mu \text{C}/{\text{m}}^2\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}}{2\left(\frac{{10}^{-9}}{36\pi }{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)}{a}_z\\ =\left(1696.46\times {10}^3\text{N}/\text{C}\times \frac{{10}^{-3}\text{kN}/\text{C}}{1\text{N}/\text{C}}\right)a_z\\ =1696.46a_z\text{kN}/\text{C}\end{array}$

Calculate the electric field $\left(E_Q\right)$ at point $\left(3,-4,10\right)$ using the relation.

  $E_P=E_3+E_{-4}+E_{10}$

  $E_P=565.486a_x\text{kN}/\text{C}+1130.97a_y\text{kN}/\text{C}+1696.46a_z\text{kN}/\text{C}$

Thus, the electric field at $Q\left(3,-4,10\right)$ is $\left(565.486a_x+1130.97a_y+1696.46a_z\right)\text{kN}/\text{C}$.