Chapter 15, Problem 15.109P

(a)

To determine

The force in the rest frame $S^{\prime }$, then transform the force to $S$ frame.

Answer to Problem 15.109P

The force in the rest frame $S^{\prime }$ is $\underset{\_}{F^{\prime }=\left(0,\frac{k{q}^2}{{r^{\prime }}^2},0\right)}$, and the force in $S$ frame is $\underset{\_}{F=\left(0,\frac{k{q}^2}{\gamma r^2},0\right)}$.

Explanation of Solution

Consider that two charges are moving in $xy$ plane. Since the charges are positioned side by side the force between them is along $y$ direction alone. The distance between the charges in the $y^{\prime }$ direction is $r^{\prime }$.

The force between the charges in its rest frame in $x$ and $y$ direction is zero $\left(F_1{}^{\prime }=0\text{and }{F^{\prime }}_2=0\right)$ respectively, and the force between the charges along $y$ direction is ${F^{\prime }}_2=\frac{k{q}^2}{{r^{\prime }}^2}$

Write the net force between the two charges in $S^{\prime }$ frame.

    $F^{\prime }=\left(0,\frac{k{q}^2}{{r^{\prime }}^2},0\right)$        (I)

But $F_1{}^{\prime }=\frac{F_1-\beta F\cdot \frac{v}{c}}{1-\beta \frac{v_1}{c}}$, ${F^{\prime }}_2=\frac{F_2}{\gamma \left(1-\beta \frac{v_1}{c}\right)}$, and ${F^{\prime }}_3=\frac{F_2}{\gamma \left(1-\beta \frac{v_1}{c}\right)}$

Using $F_1{}^{\prime }$, write the inverse transformation for $F_1{}^{\prime }$

    $F_1=\frac{{F^{\prime }}_1+\beta F^{\prime }\cdot \frac{v^{\prime }}{c}}{1+\beta \frac{{v^{\prime }}_1}{c}}$        (II)

Using $F_2{}^{\prime }$, write the inverse transformation for $F_2{}^{\prime }$

    $F_2=\frac{{F^{\prime }}_2}{\gamma \left(1+\beta \frac{{v^{\prime }}_1}{c}\right)}$        (III)

Using $F_3{}^{\prime }$, write the inverse transformation for $F_3{}^{\prime }$

    $F_3=\frac{{F^{\prime }}_3}{\gamma \left(1+\beta \frac{{v^{\prime }}_1}{c}\right)}$        (IV)

Since ${\text{V}}^{\prime }=\left({v^{\prime }}_1,0,0\right)$, therefore $F^{\prime }\cdot v^{\prime }=0$.

Use the above equation in equation (II).

    $F_1=0$        (V)

Since ${\text{V}}^{\prime }=\left({v^{\prime }}_1,0,0\right)$, therefore $F^{\prime }\cdot v^{\prime }=0$.

Use the above equation in equation (IV).

    $F_3=0$        (VI)

Write the equation for net force between the charges in $S$ frame.

    $F=\left(0,\frac{k{q}^2}{\gamma r^2},0\right)$

Conclusion:

Therefore, the force in the rest frame $S^{\prime }$ is $\underset{\_}{F^{\prime }=\left(0,\frac{k{q}^2}{{r^{\prime }}^2},0\right)}$, and the force in $S$ frame is $\underset{\_}{F=\left(0,\frac{k{q}^2}{\gamma r^2},0\right)}$.

(b)

To determine

The electric field and magnetic field in $S^{\prime }$ frame and in $S$ frame using the transformation equation, also compute the Lorentz force on either charge in $S$ frame.

Answer to Problem 15.109P

The electric field in $S^{\prime }$ frame is $\underset{\_}{{\text{E}}^{\prime }\text{=}\left(0,\frac{k{q}^2}{{r^{\prime }}^2},0\right)}$ and magnetic field in $S^{\prime }$ frame is $\underset{\_}{{\text{B}}^{\prime }=\left(0,0,0\right)}$. The electric field in $S$ frame is $\underset{\_}{\text{E=}\left(0,\frac{\gamma k{q}^2}{r^2},0\right)}$ and magnetic field in $S$ frame is $\underset{\_}{\text{B}=\left(0,0,\frac{\gamma \beta kq}{c{r}^2}\right)}$. The Lorentz force on either charge in $S$ frame is $\underset{\_}{\text{F}=\frac{\gamma k{q}^2}{r^2}\left[1-{\beta }^2\right]j}$.

Explanation of Solution

In $S^{\prime }$ frame, electric field at one charge due to the other is given by

    $E^{\prime }=\frac{F^{\prime }}{q}$

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

Write the expression for $E_1{}^{\prime }$.

    $E_1{}^{\prime }=\frac{F_1{}^{\prime }}{q}$

Since ${F^{\prime }}_1=0$, the above equation is written as

    ${E^{\prime }}_1=0$        (VII)

Write the expression for $E_2{}^{\prime }$.

    $E_2{}^{\prime }=\frac{F_2{}^{\prime }}{q}$

Since $F_2{}^{\prime }=\frac{k{q}^2}{{r^{\prime }}^2}$, the above equation is written as

    $\begin{array}{c}{E}_2{}^{\prime }=\frac{\left(\frac{k{q}^2}{{r^{\prime }}^2}\right)}{q}\\ =\frac{kq}{{r^{\prime }}^2}\end{array}$        (VIII)

Write the expression for $E_3{}^{\prime }$.

    $E_3{}^{\prime }=\frac{F_3{}^{\prime }}{q}$

Since ${F^{\prime }}_3=0$, the above equation is written as

    ${E^{\prime }}_3=0$        (IX)

The net electric field in $S^{\prime }$ frame is given by

    ${\text{E}}^{\prime }=\left(0,\frac{kq}{{r^{\prime }}^2},0\right)$        (X)

Both the charges are moving with the same speed, thus the relative speed of both the charges is zero. Since each charge is stationary they produce zero magnetic field $\left({\text{B}}^{\prime }\text{=}\left(0,0,0\right)\right)$.

From transformation equation, ${E^{\prime }}_1$ is given by

    ${E^{\prime }}_1=E_1$        (XI)

From the above equation $E_1=0$, since ${E^{\prime }}_1=0$.

From transformation equation, ${B^{\prime }}_1$ is given by

    ${B^{\prime }}_1=B_1$        (XII)

From the above equation $B_1=0$, since ${B^{\prime }}_1=0$.

From transformation equation, ${E^{\prime }}_3$ is given by

    ${E^{\prime }}_3=\gamma \left(E_3+\beta c{B}_2\right)$        (XIII)

Since ${E^{\prime }}_3=0$, the above equation can be written as

    $\gamma \left(E_3+\beta c{B}_2\right)=0$        (XIV)

From transformation equation, ${B^{\prime }}_2$ is given by

    ${B^{\prime }}_2=\gamma \left(B_2+\beta \frac{E_3}{c}\right)$        (XV)

Since $B_2{}^{\prime }=0$, the above equation can be written as

    $\gamma \left(B_2+\beta \frac{E_3}{c}\right)=0$        (XVI)

By solving equation (XIV) and (XV), we conclude that $E_3=B_2=0$.

From transformation equation, ${E^{\prime }}_2$ is given by

    ${E^{\prime }}_2=\gamma \left(E_2-\beta c{B}_3\right)$        (XVII)

Since $E_2{}^{\prime }=\frac{kq}{{r^{\prime }}^2}$, the above equation is written as

    $\begin{array}{c}\frac{kq}{{r^{\prime }}^2}=\gamma \left(E_2-\beta c{B}_3\right)\\ E_2-\beta c{B}_3=\frac{kq}{\gamma {r^{\prime }}^2}\end{array}$        (XVIII)

From transformation equation, ${B^{\prime }}_3$ is given by

    ${B^{\prime }}_3=\gamma \left(B_3-\beta \frac{E_2}{c}\right)$        (XIX)

Since ${B^{\prime }}_3=0$, the above equation is written as

    $\begin{array}{c}\gamma \left(B_3-\beta \frac{E_2}{c}\right)=0\\ B_3=\beta \frac{E_2}{c}\end{array}$        (XX)

Use the above equation in equation (XVIII).

    $\begin{array}{c}{E}_2-\beta c\left(\beta \frac{E_2}{c}\right)=\frac{kq}{\gamma {r^{\prime }}^2}\\ E_2\left(1-{\beta }^2\right)=\frac{kq}{\gamma {r^{\prime }}^2}\\ E_2\left(\frac{1}{{\gamma }^2}\right)=\frac{kq}{\gamma {r^{\prime }}^2}\\ E_2=\frac{\gamma kq}{{r^{\prime }}^2}\end{array}$

Since $r\text{'}=r$, the above equation is written as

    $E_2=\frac{\gamma kq}{r^2}$        (XXI)

Use equation (XXI) in (XX).

    $\begin{array}{c}{B}_3=\beta \frac{\left(\frac{\gamma kq}{r^2}\right)}{c}\\ B_3=\frac{\gamma \beta kq}{c{r}^2}\end{array}$        (XXII)

The net electric field in $S$ frame is derived as

    $\text{E=}\left(0,\frac{\gamma kq}{r^2},0\right)$        (XXIII)

The net magnetic field in $S$ frame is derived as

    $\text{B}=\left(0,0,\frac{\gamma \beta kq}{c{r}^2}\right)$        (XXIV)

Write the Lorentz force on each charge in $S$ frame.

    $\text{F}=q\left(\text{E}+\text{V}\times \text{B}\right)$

Here, $\text{F}$ is the force on each charge, $q$ is the charge, $\text{E}$ is the electric field, $\text{V}$ is the velocity, and $\text{B}$ is the magnetic field.

Use equation (XXIII) and (XXIV) in the above equation.

    $\begin{array}{c}\text{F}=q\left[\frac{\gamma kq}{r^2}j+\left(vi\times \frac{\gamma \beta kq}{c{r}^2}k\right)\right]\\ =\left[\frac{\gamma k{q}^2}{r^2}-\frac{v\gamma \beta k{q}^2}{c{r}^2}\right]j\\ =\frac{\gamma k{q}^2}{r^2}\left[1-\left(\frac{v}{c}\right)\beta \right]j\\ =\frac{\gamma k{q}^2}{r^2}\left[1-{\beta }^2\right]j\end{array}$        (XXV)

The electrostatic force is positive thus it is attractive, and the magnetic force is negative thus it is responsive.

When $\beta$ tends to $1$, the magnetic force becomes $\frac{\gamma k{q}^2}{r^2}$ equal to the electrostatic force. And also the force $\text{F}$ equals to zero. Thus both the electric and magnetic force cancels each other.

Conclusion:

Therefore, the electric field in $S^{\prime }$ frame is $\underset{\_}{{\text{E}}^{\prime }\text{=}\left(0,\frac{k{q}^2}{{r^{\prime }}^2},0\right)}$ and magnetic field in $S^{\prime }$ frame is $\underset{\_}{{\text{B}}^{\prime }=\left(0,0,0\right)}$. The electric field in $S$ frame is $\underset{\_}{\text{E=}\left(0,\frac{\gamma k{q}^2}{r^2},0\right)}$ and magnetic field in $S$ frame is $\underset{\_}{\text{B}=\left(0,0,\frac{\gamma \beta kq}{c{r}^2}\right)}$. The Lorentz force on either charge in $S$ frame is $\underset{\_}{\text{F}=\frac{\gamma k{q}^2}{r^2}\left[1-{\beta }^2\right]j}$.