12.46 Two charges, $\pm q$, are on parallel trajectories a distance apart, moving with equal speeds in opposite directions. We’re interested in the force on$+q$ due to$-q$ at the instant they cross (Fig. 12.42). Fill in the following table, doing all the consistency checks you can think of as you go alon

	System (Fig. 12.42)	System ( at rest)	System ( at rest)
$E$at $+q$due to $-q$
localid="1658130749562" $B$at$+q$due to $-q$
$F$at $+q$due to$-q$

The expression for the force from the Coulomb’s law is given by;

$F=\frac{1}{4\pi {\epsilon }_o}\frac{q_1{q}_2}{r^2}$

Here$F$ is the magnitude of force between two charge,$q_1$ ,$q_2$ are the magnitude of the charges and $r$is the distance between the two charge.

The expression for the magnetic field is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\left(\frac{Q(\overrightarrow{v}\times \overrightarrow{r})}{r^2}\right)$

Here$\overrightarrow{B}$ is the magnetic field, $\overrightarrow{v}$is the velocity of the charge,$Q$ is the magnitude of the charge,$\overrightarrow{r}$ is the position vector of the charge and${\mu }_o$ is the permeability of the free space.

For the given system the position of the charge $+q$is $-\frac{d}{2}\stackrel{\wedge }{y}$and position of $-q$is $\frac{d}{2}\stackrel{\wedge }{y}$.

Electric field

The general expression for the electric field is,

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{Q}{r^2}\stackrel{\wedge }{r}$

System A;

Substitute $-q$for $Q$,$\stackrel{\wedge }{y}$for $\stackrel{\wedge }{r}$and $d$for$r$in the above equation.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$

The motion of the charges will not affect the electric field at position of $+q$due to $-q$.

System B;

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$

System C;

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$

Thus, for systems, electric field at $+q$due to is $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$.

Magnetic field

The general expression for the magnetic field is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\left(\frac{Q(\overrightarrow{v}\times \overrightarrow{r})}{r^2}\right)$

System A;

Substitute $-q$for $Q$, $\stackrel{\wedge }{y}$for $\stackrel{\wedge }{r}$,$v(-\stackrel{\wedge }{x})$ for $\overrightarrow{v}$and $d$for $r$in the above equation.

role="math" localid="1658131922476" $\begin{array}{c}\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\left(\frac{-q(v(-\stackrel{\wedge }{x})\times \stackrel{\wedge }{y})}{d^2}\right)\\ =\frac{{\mu }_o}{4\pi }\left(\frac{-qv(-\stackrel{\wedge }{z})}{d^2}\right)\\ =\frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}\end{array}$

System B;

Now if the charge $-q$is moving and$+q$is at rest, the magnetic field at the position of charge$+q$is produced due to the motion of charge $-q$.

The magnetic field produced at the position of will be same as calculated in system A.

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}$

System C;

As the charge$-q$is at rest and$+q$is at moving. The magnetic field at position of charge $+q$is zero as magnetic field is produced due to the motion of charge $-q$

$\overrightarrow{B}=0$

Thus magnetic field at system A and B is $\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}$and at system C is $\overrightarrow{B}=0$.

Force

The expression for the force using Lorentz law is given by,

$\overrightarrow{F}=Q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$...... (1)

System A;

Substitute$+q$for$Q$, $\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$for$\overrightarrow{E}$,$-v\left(\stackrel{\wedge }{x}\right)$for$\overrightarrow{v}$ and $\frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}$for $\overrightarrow{B}$in the equation (1).

$\begin{array}{c}\overrightarrow{F}=q\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}\right)+-v\left(\stackrel{\wedge }{x}\right)\times \frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}\\ =\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q^2}{d^2}\stackrel{\wedge }{y}\right)+\left(\frac{{\mu }_o}{4\pi }\left(\frac{q^2{v}^2}{d^2}\stackrel{\wedge }{y}\right)\right)\end{array}$

System B;

The velocity of the charge $+q$is zero.

Substitute $+q$for $Q$,$\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$for $\overrightarrow{E}$,$0$for $\overrightarrow{v}$and $\frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}$for$\overrightarrow{B}$in the equation (1).

role="math" localid="1658133092641" $\begin{array}{c}\overrightarrow{F}=q\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}\right)+-0\times \frac{{\mu }_o}{4\pi }\left(\frac{qv}{d^2}\right)\stackrel{\wedge }{z}\\ =\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q^2}{d^2}\stackrel{\wedge }{y}\right)\end{array}$

System C;

The velocity of the charge$-q$is zero.

Substitute $+q$for $Q$, $\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}$for $\overrightarrow{E}$, $v(-\stackrel{\wedge }{x})$for $\overrightarrow{v}$ and for $\overrightarrow{B}$in the equation (1).

.$\begin{array}{c}\overrightarrow{F}=q\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q}{d^2}\stackrel{\wedge }{y}\right)+v(-\stackrel{\wedge }{x})\times 0\\ =\left(\frac{1}{4\pi {\epsilon }_o}\frac{-q^2}{d^2}\stackrel{\wedge }{y}\right)\end{array}$

Therefore,