Chapter 1, Problem 1.19P

To determine

(a)

The field into cylindrical coordinates.

Answer to Problem 1.19P

The field into cylindrical coordinates is $A{\left({\rho }^2+z^2\right)}^{-\frac{3}{2}}\left(\rho a_{\rho }+z{a}_z\right)$.

Explanation of Solution

Concept used:

Write the expression for the $x$ -component of the rectangular system in term of cylindrical system.

   $x=\rho \cos \varphi$

Here $\rho$ is the radius of the cylindrical coordinate system and $x$ is the coordinate of the $x$ -axis.

Write the expression for the $y$ -component of the rectangular system in term of cylindrical system.

   $y=\rho \sin \varphi$

Here, $\theta$ is the angle between $z$ -axis and the line drawn from the origin and $y$ is the coordinate of the $y$ -axis.

Write the expression for the $z$ -component of the rectangular system in term of cylindrical system.

   $z=z$

Here, $z$ is the coordinate of the $z$ -axis

Calculation:

The expression for the field into rectangular coordinates is $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$.

The expression for the field in the $\rho$ coordinate system can be written as.

   $F_{\rho }=F\cdot a_{\rho }$

Substitute $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$ for $F$ in the above equation.

   $\begin{array}{c}{F}_x=A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]\cdot a_{\rho }\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\left(a_x\cdot a_{\rho }\right)+y\left(a_y\cdot a_{\rho }\right)+z\left(a_z\cdot a_{\rho }\right)\right]\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\cos \varphi +y\sin \varphi +z\left(0\right)\right]\left\{\begin{array}{l}\because a_x\cdot a_{\rho }=\cos \varphi \\ \because a_y\cdot a_{\rho }=\sin \varphi \\ \because a_z\cdot a_{\rho }=0\end{array}\right\}\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\cos \varphi +y\sin \varphi \right]\end{array}$

Here, $\theta$ is the angle between $z$ -axis and the line drawn from the origin and $\varphi$ is the angle between $x$ -axis and the line drawn from the origin.

The expression for the field in the $\varphi$ coordinate system can be written as.

   $F_{\varphi }=F\cdot a_{\varphi }$

Substitute $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$ for $F$ in the above equation.

   $\begin{array}{c}{F}_{\varphi }=A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]\cdot a_{\varphi }\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\left(a_x\cdot a_{\varphi }\right)+y\left(a_y\cdot a_{\varphi }\right)+z\left(a_z\cdot a_{\varphi }\right)\right]\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[-x\sin \varphi +y\cos \varphi +z\left(0\right)\right]\left\{\begin{array}{l}\because a_x\cdot a_{\rho }=-\sin \varphi \\ \because a_y\cdot a_{\rho }=\cos \varphi \\ \because a_z\cdot a_{\rho }=0\end{array}\right\}\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[-x\sin \varphi +y\cos \varphi \right]\end{array}$

The expression for the field in the $z$ coordinate system can be written as.

   $F_z=F\cdot a_z$

Substitute $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$ for $F$ in the above equation.

   $\begin{array}{c}{F}_z=A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]\cdot a_z\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\left(a_x\cdot a_z\right)+y\left(a_y\cdot a_z\right)+z\left(a_z\cdot a_z\right)\right]\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\left(0\right)+y\left(0\right)+z\left(1\right)\right]\left\{\begin{array}{l}\because a_x\cdot a_z=0\\ \because a_y\cdot a_z=0\\ \because a_z\cdot a_z=1\end{array}\right\}\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[z\right]\end{array}$

The expression for the field in of the cylindrical coordinate system can be written as the sum of the all three component of the cylindrical coordinate system.

   $\begin{array}{c}F=\left\{\begin{array}{l}A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x\cos \varphi +y\sin \varphi \right]\\ +A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[-x\sin \varphi +y\cos \varphi \right]\\ +A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[z\right]\end{array}\right\}\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left(x\cos \varphi +y\sin \varphi -x\sin \varphi +y\cos \varphi +z\right)\end{array}$

Substitute $\rho \cos \varphi$ for $x$ , $\rho \sin \varphi$ for $y$ and $z$ for $z$ in the above equation.

   $\begin{array}{c}F=A{\left({\left(\rho \cos \varphi \right)}^2+{\left(\rho \sin \varphi \right)}^2+z^2\right)}^{-\frac{3}{2}}\left(\begin{array}{l}\left(\left(\rho \cos \varphi \right)\cos \varphi +\left(\rho \sin \varphi \right)\sin \varphi \right)a_{\rho }\\ +\left(-\left(\rho \cos \varphi \right)\sin \varphi +\left(\rho \sin \varphi \right)\cos \varphi \right)a_{\varphi }+z{a}_z\end{array}\right)\\ =A{\left({\rho }^2{\cos }^2\varphi +{\rho }^2{\sin }^2\varphi +z^2\right)}^{-\frac{3}{2}}\left(\begin{array}{l}\left(\rho {\cos }^2\varphi +\rho {\sin }^2\varphi \right)a_{\rho }\\ +\left(-\rho \cos \varphi \sin \varphi +\rho \sin \varphi \cos \varphi \right)a_{\varphi }+z{a}_z\end{array}\right)\\ =A{\left({\rho }^2{\cos }^2\varphi +{\rho }^2{\sin }^2\varphi +z^2\right)}^{-\frac{3}{2}}\left(\rho \left({\cos }^2\varphi +{\sin }^2\varphi \right)a_{\rho }+z{a}_z\right)\\ =A{\left({\rho }^2\left({\cos }^2\varphi +{\sin }^2\varphi \right)+z^2\right)}^{-\frac{3}{2}}\left(\rho \left(1\right)a_{\rho }+z{a}_z\right)\end{array}$

Simplify further.

   $\begin{array}{c}F=A{\left({\rho }^2\left(1\right)+z^2\right)}^{-\frac{3}{2}}\left(\rho a_{\rho }+z{a}_z\right)\\ =A{\left({\rho }^2+z^2\right)}^{-\frac{3}{2}}\left(\rho a_{\rho }+z{a}_z\right)\end{array}$

Conclusion:

Thus, the field into cylindrical coordinates is $A{\left({\rho }^2+z^2\right)}^{-\frac{3}{2}}\left(\rho a_{\rho }+z{a}_z\right)$.

To determine

(b)

The field into rectangular coordinates.

Answer to Problem 1.19P

The expression for the field into rectangular coordinates is $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$.

Explanation of Solution

Given:

The expression for the field in spherical coordinate $F$ is $A{r}^{-2}{a}_r$.

Concept used:

Write the expression for the radius of the spherical system in term of rectangular system.

   $r=\sqrt{x^2+y^2+z^2}$

Here $r$ is the radius of the spherical coordinate system.

Write the expression for the angle between $z$ -axis and the line drawn from the origin in term of rectangular system.

   $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

Write the expression for the angle between $x$ -axis and the line drawn from the origin in term of rectangular system.

   $\varphi ={\tan }^{-1}\frac{y}{x}$

Calculation:

The expression for $\cos \theta$ can be can be written as.

   $\cos \theta =\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

By Pythagoras theorem $\sin \theta$ can be written as.

   $\sin \theta =\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$

By Pythagoras theorem $\sin \varphi$ can be written as.

   $\sin \varphi =\frac{y}{\sqrt{x^2+y^2}}$

By Pythagoras theorem $\cos \varphi$ can be written as.

   $\cos \varphi =\frac{x}{\sqrt{x^2+y^2}}$

The expression for the field in the $x$ coordinate system can be written as.

   $F_x=F\cdot a_x$

Here, $a_x$ is the unit vector of the $x$ component of the rectangular coordinate.

Substitute $A{r}^{-2}{a}_r$ for $F$ in the above equation.

   $\begin{array}{c}{F}_x=A{r}^{-2}{a}_r\cdot a_x\\ =A{r}^{-2}\left(a_r\cdot a_x\right)\text{}\\ =A{r}^{-2}\left(\sin \theta \cos \varphi \right)\left\{\because a_x\cdot a_r=\sin \theta \cos \varphi \right\}\end{array}$

The expression for the field in the $y$ coordinate system can be written as.

   $F_x=F\cdot a_y$

Here, $a_y$ is the unit vector of the $y$ -component of the rectangular coordinate system.

Substitute $A{r}^{-2}{a}_r$ for $F$ in the above equation.

   $\begin{array}{c}{F}_y=A{r}^{-2}{a}_r\cdot a_y\\ =A{r}^{-2}\left(a_r\cdot a_y\right)\\ =A{r}^{-2}\left(\sin \theta \sin \varphi \right)\left\{\because a_r\cdot a_y=\sin \theta \sin \varphi \right\}\end{array}$

The expression for the field in the $z$ coordinate system can be written as.

   $F_z=F\cdot a_z$

Here, $a_z$ is the unit vector of the $z$ -component of the rectangular coordinate system.

Substitute $A{r}^{-2}{a}_r$ for $F$ in the above equation.

   $\begin{array}{c}{F}_z=A{r}^{-2}{a}_r\cdot a_z\\ =A{r}^{-2}\left(a_r\cdot a_z\right)\\ =A{r}^{-2}\left(\cos \theta \right)\left\{\because a_r\cdot a_z=\cos \theta \right\}\end{array}$

The expression for the field in of the rectangular coordinate system can be written as the sum of the all three component of the rectangular coordinate system.

   $\begin{array}{c}F=A{r}^{-2}\left(\sin \theta \cos \varphi \right)a_x+A{r}^{-2}\left(\sin \theta \sin \varphi \right)a_y+A{r}^{-2}\left(\cos \theta \right)a_z\\ =A{r}^{-2}\left[\left(\sin \theta \cos \varphi \right)a_x+\left(\sin \theta \sin \varphi \right)a_y+\left(\cos \theta \right)a_z\right]\end{array}$

Substitute $\sqrt{x^2+y^2+z^2}$ for $r$ , $\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$ for $\sin \theta$ , $\frac{x}{\sqrt{x^2+y^2}}$ for $\cos \varphi$ , $\frac{y}{\sqrt{x^2+y^2}}$ for $\sin \varphi$ and $\frac{z}{\sqrt{x^2+y^2+z^2}}$ for $\cos \theta$ in the above equation.

   $\begin{array}{c}F=A{\left(\sqrt{x^2+y^2+z^2}\right)}^{-2}\left[\begin{array}{c}\left(\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right)\right)a_x+\\ \left(\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\left(\frac{y}{\sqrt{x^2+y^2}}\right)\right)a_y+\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)a_z\end{array}\right]\\ =A\frac{{\left(\sqrt{x^2+y^2+z^2}\right)}^{-2}}{\sqrt{x^2+y^2+z^2}}\left[x{a}_x+y{a}_y+z{a}_z\right]\\ =A\frac{1}{\left(\sqrt{x^2+y^2+z^2}\right){\left(\sqrt{x^2+y^2+z^2}\right)}^2}\left[x{a}_x+y{a}_y+z{a}_z\right]\\ =A\frac{1}{\left(\sqrt{x^2+y^2+z^2}\right)\left(x^2+y^2+z^2\right)}\left[x{a}_x+y{a}_y+z{a}_z\right]\end{array}$

Simplify further.

   $\begin{array}{c}F=A\frac{1}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\left[x{a}_x+y{a}_y+z{a}_z\right]\\ =A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]\end{array}$

Conclusion:

Thus, the expression for the field into rectangular coordinates is

   $A{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left[x{a}_x+y{a}_y+z{a}_z\right]$.