Question: Suppose a point charge q is constrained to move along the x axis. Show that the fields at points on the axis to the right of the charge are given by

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{B}\mathbf{=}\mathbf{0}$

(Do not assume is constant!) What are the fields on the axis to the left of the charge?

The point charge moves along the axis.

The expression for electric field due to moving charge is given as follows.

$\mathit{E}\left(r,t\right)\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\left(r\times u\right)}^{\mathbf{3}}}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right]$ …… (1)

Here, is the charge, is the speed, and is the velocity of light.

The expression for the magnetic field strength is given as follows.

$\mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}\left(\hat{r}\times E\right)$ …… (2)

The electric field due to moving point charge is given by,

$E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right]$ …… (3)

Here particle in constrained to moves along the axis.

Therefore, the points on the right side of the charge,

localid="1657878176173" $$\begin{gathered} v=v\hat{x}\backslash \\ a=a\hat{x} \\ \hat{r}=\hat{x} \\ r·u=r\left(c-v\right) \end{gathered}$$

Since,

$$\begin{gathered} u=\left(c-v\right)\hat{x} \\ r=r\hat{x} \\ u\times a=\left(c-v\right)\hat{x}\times a\hat{x} \\ u\times a=0 \end{gathered}$$

Recall equation (3),

$E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right]$

Substitute $\left(c-v\right)\hat{x}$for u,0, $r\left(c-v\right)$for r.u for into above equation.

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r\left(c-v\right)\right)}^3}\left[\left(c^2-v^2\right)\left(c-v\right)\hat{x}+r\times \left(0\right)\right] \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{r^3{\left(c-v\right)}^3}\left[\left(c+v\right)\left(c-v\right)\left(c-v\right)\hat{x}\right] \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2{\left(c-v\right)}^3}\left[\left(c+v\right){\left(c-v\right)}^2\hat{x}\right] \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}\backslash \end{gathered}$$

Calculate the expression for magnetic field,

Substitute $\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}$for $E\left(r,t\right)$and for into equation (3).

$$\begin{gathered} B=\frac{1}{c}\left(\hat{x}\times \frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}\right) \\ B=0 \end{gathered}$$

The points on the right side of the charge,

$$\begin{gathered} \hat{r}=-\hat{x} \\ u=-\left(c-v\right)\hat{x}\backslash \\ r·u=\left(c+v\right)r\backslash \\ u\times a=0 \end{gathered}$$

Recall equation (3),

$E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right]$

Substitute $-\left(c-v\right)\hat{x}$for ,u,0 for uxa and $r\left(c+v\right)$ for r.u into above equation.

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r\left(c+v\right)\right)}^3}\left[-\left(c^2-v^2\right)\left(c+v\right)\hat{x}+r\times \left(0\right)\right] \\ E\left(r,t\right)=-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2{\left(c+v\right)}^3}\left[\left(c+v\right)\left(c-v\right)\left(c+v\right)\hat{x}\right] \\ E\left(r,t\right)=-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2{\left(c+v\right)}^3}\left[{\left(c+v\right)}^2\left(c-v\right)\hat{x}\right] \\ E\left(r,t\right)=-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x} \end{gathered}$$

Calculate the expression for magnetic field,

Substitute$-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x}$ for$E\left(r,t\right)$and for into equation (3).

$$\begin{gathered} B=\frac{1}{c}\left(-\hat{x}\times \left(-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\right)\hat{x}\right) \\ B=0 \end{gathered}$$

Hence the expression for electric and magnetic field on the right side is showed and the electric and magnetic field on left side are$-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x}$ and 0 respectively.