Chapter 15, Problem 15.110P

(a)

To determine

The electric field and magnetic field in the charge’s rest frame $S^{\prime }$.

Answer to Problem 15.110P

The electric field in the charge’s rest frame $S^{\prime }$ is $\underset{\_}{{\text{E}}^{\prime }\text{=}\frac{kq\text{<msup><mover accent="true">r<mo>^</mo></mover><mo>′</mo></msup>}}{{r^{\prime }}^2}}$ and magnetic field in the charge’s rest frame $S^{\prime }$ is $\underset{\_}{{\text{B}}^{\prime }=0}$.

Explanation of Solution

Figure 1 represents the point P with position vector $\text{r}$ in $S$ frame, and the vector pointing from charge position to point P is $\text{R}$ which is given by $\text{R}=\text{r}-vt\widehat{\text{x}}$. The charge $q$ is moving with a speed $v$ along the $x$ axis with a speed $v$ in $S$ frame.

Since the charge is in rest frame $S^{\prime }$ it produces only electric field, and the magnetic field is zero.

Write the equation for ${\text{E}}^{\prime }$.

    ${\text{E}}^{\prime }\text{=}\frac{kq\text{<msup><mover accent="true">r<mo>^</mo></mover><mo>′</mo></msup>}}{{r^{\prime }}^2}$

Here, ${\text{E}}^{\prime }$ is the net electric field in $S^{\prime }$ frame.

Since the net electric field in $S^{\prime }$ frame is ${\text{E}}^{\prime }\text{=}\frac{kq\text{<msup><mover accent="true">r<mo>^</mo></mover><mo>′</mo></msup>}}{{r^{\prime }}^2}$, its component along $x,y,\text{and z}$ is given by ${E^{\prime }}_x=\frac{kq{x}^{\prime }}{{r^{\prime }}^3}$, ${E^{\prime }}_y=\frac{kq{y}^{\prime }}{{r^{\prime }}^3}$, and ${E^{\prime }}_z=\frac{kq{z}^{\prime }}{{r^{\prime }}^3}$ respectively.

Since the net electric field in $S^{\prime }$ frame is ${\text{B}}^{\prime }\text{=}0$, its component along $x,y,\text{and z}$ is given by ${B^{\prime }}_x=0$, ${B^{\prime }}_y=0$, and ${B^{\prime }}_z=0$ respectively.

Conclusion:

Therefore, the electric field in the charge’s rest frame $S^{\prime }$ is $\underset{\_}{{\text{E}}^{\prime }\text{=}\frac{kq\text{<msup><mover accent="true">r<mo>^</mo></mover><mo>′</mo></msup>}}{{r^{\prime }}^2}}$ and magnetic field in the charge’s rest frame $S^{\prime }$ is $\underset{\_}{{\text{B}}^{\prime }=0}$.

(b)

To determine

To show that electric field as $\text{E}=\frac{kq\left(1-{\beta }^2\right)}{{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\frac{\widehat{\text{R}}}{R^2}$.

Answer to Problem 15.110P

It is shown that electric field $\underset{\_}{\text{E}=\frac{kq\left(1-{\beta }^2\right)}{{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\frac{\widehat{\text{R}}}{R^2}}$.

Explanation of Solution

Write the expression for ${E^{\prime }}_x$.

    ${E^{\prime }}_x=E_x$        (I)

Here, ${E^{\prime }}_x$ is the electric field along $x$ direction in $S^{\prime }$ frame, and $E_x$ is the original electric field component along $x$ direction in $S$ frame.

Write the expression for ${E^{\prime }}_y$.

    ${E^{\prime }}_y=\gamma \left(E_y-\beta B_z\right)$        (II)

Here, ${E^{\prime }}_y$ is the electric field along $y$ direction in $S^{\prime }$ frame.

Rearrange the above equation, to get $E_y$.

    $E_y=\frac{{E^{\prime }}_y}{\gamma }+\beta c{B}_z$

Here, $E_y$ is the electric field along $y$ direction in $S$ frame.

Since $B_z=\frac{{B^{\prime }}_z}{\gamma }+\beta \frac{E_y}{c}$, the above equation is written as

    $E_y=\frac{{E^{\prime }}_y}{\gamma }+\beta c\left(\frac{{B^{\prime }}_z}{\gamma }+\beta \frac{E_y}{c}\right)$        (III)

Write the expression for ${E^{\prime }}_z$.

    ${E^{\prime }}_z=\gamma \left(E_z+\beta c{B}_y\right)$        (IV)

Here, ${E^{\prime }}_z$ is the electric field along $z$ direction in $S^{\prime }$ frame.

Rearrange the above equation to get $E_z$.

    $E_z=\frac{{E^{\prime }}_z}{\gamma }-\beta c\gamma \left({B^{\prime }}_y-\beta {E^{\prime }}_z/c\right)$

Here, $E_z$ is the electric field along $z$ direction in $S$ frame.

Since $\gamma \left({B^{\prime }}_y-\beta {E^{\prime }}_z/c\right)=B_y$ the above equation is reduced as

    $E_z=\frac{{E^{\prime }}_z}{\gamma }-\beta c{B}_y$        (V)

Write the expression for $B_x{}^{\prime }$.

    ${B^{\prime }}_x=B_x$        (VI)

Here, ${B^{\prime }}_x$ is the magnetic field component along $x$ direction in $S^{\prime }$ frame.

Write the expression $B_y{}^{\prime }$.

    ${B^{\prime }}_y=\gamma \left(B_y+\frac{\beta E_z}{c}\right)$        (VII)

Here, ${B^{\prime }}_y$ is the magnetic field component along $y$ direction in $S^{\prime }$ frame.

Rearrange the above equation to get $B_y$.

    $B_y=\gamma \left({B^{\prime }}_y-\beta {E^{\prime }}_z/c\right)$        (VIII)

Here, $B_y$ is the magnetic field component along $y$ direction in $S$ frame.

Write the expression for ${B^{\prime }}_z$.

    ${B^{\prime }}_z=\gamma \left(B_z-\beta E_y/c\right)$        (IX)

Here, ${B^{\prime }}_z$ is the magnetic field component along $z$ direction in $S^{\prime }$ frame.

Rearrange the above equation to get $B_z$.

    $B_z=\gamma \left({B^{\prime }}_z-\beta {E^{\prime }}_y/c\right)$        (X)

Here, $B_z$ is the magnetic field component along $z$ direction in $S$ frame.

From equation (III).

    $\begin{array}{c}{E}_y=\frac{{E^{\prime }}_y}{\gamma }+\beta \left[\frac{{B^{\prime }}_z}{\gamma }+\beta \frac{E_y}{c}\right]\\ E_y=\frac{{E^{\prime }}_y}{\gamma }+\frac{\beta c{B^{\prime }}_z}{\gamma }+{\beta }^2{E}_y\\ \left(1-{\beta }^2\right)E_y=\frac{1}{\gamma }\left({E^{\prime }}_y+\beta c{B^{\prime }}_z\right)\end{array}$

Since $1-{\beta }^2=\frac{1}{{\gamma }^2}$, the above equation can be written as

    $\begin{array}{c}\frac{E_y}{{\gamma }^2}=\frac{1}{\gamma }\left({E^{\prime }}_y+\beta c{B^{\prime }}_z\right)\\ E_y=\gamma \left({E^{\prime }}_y+\beta c{B^{\prime }}_z\right)\\ E_y=\frac{{E^{\prime }}_y+\beta c{B^{\prime }}_z}{\sqrt{1-{\beta }^2}}\end{array}$

Since the value of $B_z{}^{\prime }$ is equal to zero $B_z{}^{\prime }=0$, the above equation is written as

    $E_y=\frac{{E^{\prime }}_y}{\sqrt{1-{\beta }^2}}$

Since $E_y{}^{\prime }=\frac{kq{y}^{\prime }}{{r^{\prime }}^3}$, the above equation is written as

    $E_y=\frac{1}{\sqrt{1-{\beta }^2}}\frac{kq{y}^{\prime }}{{r^{\prime }}^3}$        (XI)

From equation (VIII), $B_y$ is written as

    $\begin{array}{c}{B}_y=\frac{B_y{}^{\prime }}{\gamma }-\frac{\beta }{c}\left(\frac{E_z{}^{\prime }}{\gamma }-\beta c{B}_y\right)\\ =\frac{B_y{}^{\prime }}{\gamma }-\frac{\beta E_z{}^{\prime }}{c\gamma }+{\beta }^2{B}_y\\ \left(1-{\beta }^2\right)B_y=\frac{1}{\gamma }\left({B^{\prime }}_y-\frac{\beta }{c}{E}_z{}^{\prime }\right)\\ B_y=\gamma \left({B^{\prime }}_y-\frac{\beta }{c}{E}_z{}^{\prime }\right)\end{array}$

Since $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$, the above equation is written as

    $B_y=\frac{{B^{\prime }}_y-\frac{\beta E_z{}^{\prime }}{c}}{\sqrt{1-{\beta }^2}}$        (XII)

From equation (V), $E_z$ is written as

    $\begin{array}{c}{E}_z=\frac{{E^{\prime }}_z}{\gamma }-\beta c\gamma \left({B^{\prime }}_y-\beta {E^{\prime }}_z/c\right)\\ =\gamma \left[\left(\frac{1}{{\gamma }^2}+{\beta }^2\right){E^{\prime }}_z-\beta c{B^{\prime }}_y\right]\\ =\gamma \left[\left(1-{\beta }^2+{\beta }^2\right){E^{\prime }}_z-\beta c{B^{\prime }}_y\right]\\ =\gamma \left[{E^{\prime }}_z-\beta c{B^{\prime }}_y\right]\end{array}$

Since $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$, the above equation is written as

    $E_z=\frac{{E^{\prime }}_z-\beta c{B^{\prime }}_y}{\sqrt{1-{\beta }^2}}$

Since the value of $B_y{}^{\prime }$ is equal to zero $B_y{}^{\prime }=0$, the above equation is written as

    $E_z=\frac{{E^{\prime }}_z}{\sqrt{1-{\beta }^2}}$

Since $E_z{}^{\prime }=\frac{kq{z}^{\prime }}{{r^{\prime }}^3}$, the above equation is written as

    $E_z=\frac{1}{\sqrt{1-{\beta }^2}}\frac{kq{z}^{\prime }}{{r^{\prime }}^3}$        (XIII)

From equation (IX), $B_z$ is written as

    $\begin{array}{c}{B}_z=\frac{{B^{\prime }}_z}{\gamma }+\beta \frac{E_y}{c}\\ =\frac{{B^{\prime }}_z}{\gamma }+\frac{\beta }{c}\gamma \left({E^{\prime }}_y+\beta c{B^{\prime }}_z\right)\\ =\gamma \left[{B^{\prime }}_z\left(\frac{1}{{\gamma }^2}+{\beta }^2\right)+\beta {E^{\prime }}_y/c\right]\\ =\gamma \left[{B^{\prime }}_z+\beta {E^{\prime }}_y/c\right]\end{array}$

Since $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$, the above equation is written as

    $B_z=\frac{{B^{\prime }}_z+\beta {E^{\prime }}_y/c}{\sqrt{1-{\beta }^2}}$        (XIV)

From Lorentz transformation, $x^{\prime }=\gamma \left(x-vt\right)$, $y^{\prime }=y$, $z^{\prime }=z$, and $t^{\prime }=\gamma \left(t-\frac{vx}{c^2}\right)$

Write the equation for $R^2$.

    $R^2={\left(x-vt\right)}^2+y^2+z^2$

From Figure 1 it is known that $x-vt=R\cos \theta$, the above equation is written as

    $\begin{array}{c}{R}^2=R^2{\cos }^2\theta +y^2+z^2\\ y^2+z^2=R^2\left(1-{\cos }^2\theta \right)\\ y^2+z^2=R^2\left({\sin }^2\theta \right)\end{array}$        (XV)

Write the expression for ${r^{\prime }}^2$.

    ${r^{\prime }}^2={x^{\prime }}^2+{y^{\prime }}^2+{z^{\prime }}^2$

Use equation (XV) and $x^{\prime }=\gamma \left(x-vt\right)$ in the above equation.

    ${r^{\prime }}^2={\gamma }^2{\left(x-vt\right)}^2+R^2{\sin }^2\theta$

Use $x-vt=R\cos \theta$ in the above equation.

    $\begin{array}{c}{r^{\prime }}^2=\frac{R^2{\cos }^2\theta }{1-{\beta }^2}+R^2{\sin }^2\theta \\ =\frac{R^2}{1-{\beta }^2}\left[{\cos }^2\theta +\left(1-{\beta }^2\right){\sin }^2\theta \right]\\ =\frac{R^2}{1-{\beta }^2}\left[{\cos }^2\theta +{\sin }^2\theta -{\beta }^2{\sin }^2\theta \right]\\ =\frac{R^2}{1-{\beta }^2}\left(1-{\beta }^2{\sin }^2\theta \right)\end{array}$

Rearrange the above equation.

    $r^{\prime }=\frac{R}{\sqrt{1-{\beta }^2}}{\left(1-{\beta }^2{\sin }^2\theta \right)}^{1/2}$        (XVI)

Since $E_x={E^{\prime }}_x=\frac{kq{x}^{\prime }}{r^3}$, equation (I) is written as

    $E_x=\frac{kq{x}^{\prime }}{r^3}$

Use equation (XVI), and $x^{\prime }=x$ in the above equation.

    $\begin{array}{c}{E}_x=\frac{kqx}{{\left[\frac{R}{\sqrt{1-{\beta }^2}}{\left(1-{\beta }^2{\sin }^2\theta \right)}^{1/2}\right]}^3}\\ E_x=\frac{kq(x-vt)\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\end{array}$        (XVII)

Use equation (XVI), and $y^{\prime }=y$ in the equation (XI).

    $\begin{array}{c}{E}_y=\frac{1}{\sqrt{1-{\beta }^2}}\frac{kqy}{{\left[\frac{R}{\sqrt{1-{\beta }^2}}{\left(1-{\beta }^2{\sin }^2\theta \right)}^{1/2}\right]}^3}\\ E_y=\frac{kqy\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\end{array}$        (XVIII)

Use equation (XVI), and $z^{\prime }=z$ in the equation (XIII)

    $\begin{array}{c}{E}_z=\frac{1}{\sqrt{1-{\beta }^2}}\frac{kqz}{{\left[\frac{R}{\sqrt{1-{\beta }^2}}{\left(1-{\beta }^2{\sin }^2\theta \right)}^{1/2}\right]}^3}\\ E_z=\frac{kqz\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\end{array}$        (XIX)

Write the expression for $\text{E}$

    $\text{E}=E_{x}i+E_{y}j+E_{z}k$

Use equation (XVII), (XVIII), and (IX) in the above equation.

    $\begin{array}{c}\text{E=}\frac{kq(x-vt)\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}i+\frac{kqy\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}j+\frac{kqz\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}k\\ =\frac{kq\left(1-{\beta }^2\right)}{R^3{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\left[\left(x-vt\right)i+yj+zk\right]\\ =\frac{kq\left(1-{\beta }^2\right)}{{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\frac{R}{R^3}\\ =\frac{kq\left(1-{\beta }^2\right)}{\left(1-{\beta }^2{\sin }^2\theta \right)}\frac{\widehat{\text{R}}}{R^2}\end{array}$

Conclusion:

Therefore, It is shown that electric field $\underset{\_}{\text{E}=\frac{kq\left(1-{\beta }^2\right)}{{\left(1-{\beta }^2{\sin }^2\theta \right)}^{3/2}}\frac{\widehat{\text{R}}}{R^2}}$.

(c)

To determine

To sketch the behavior of field strength as a function of $\theta$.

Answer to Problem 15.110P

The behavior of field strength as a function of $\theta$ is sketched in Figure 2.

Explanation of Solution

Figure 2 represents the behavior of field strength as a function of $\theta$.

Conclusion:

Therefore, the behavior of field strength as a function of $\theta$ is sketched in Figure 2.