A uniformly charged rod (length L, charge density $\mathit{\lambda }$ ) slides out thex axis at constant speedv. At time t = 0 the back end passes the origin (so its position as a function of time is x = vt , while the front end is at x = vt + L ). Find the retarded scalar potential at the origin, as a function of time, for t > 0 . [First determine the retarded time t₁ for the back end, the retarded time t₂ for the front end, and the corresponding retarded positions x₁ and x₂ .] Is your answer consistent with the Liénard-Wiechert potential, in the point charge limit (L << vt , with $\mathit{\lambda }\mathit{L}\mathbf{=}\mathit{q}$)? Do not assume v << c .

Let x₁ be the retarded position for the back end of the rod and x₂ be the position for the front end of a rod.

Given data:

The position for the back end of a rod is x₁ = vt .

The position for the front end of a rod is x₂ = vt + L .

Write the expression for the position in terms of retarded time for the back end.

c ( t - t₁ ) = x₁

Substitute x₁ = vt in the above expression.

$$\begin{gathered} c\left(t-t_1\right)=v{t}_1 \\ t=t_1\left(1+\frac{v}{c}\right) \\ {t}_1=\frac{t}{1+\frac{v}{c}} \end{gathered}$$

Write the expression for the position in terms of retarded time for the front end.

c ( t - t₂ ) = x₂

Substitute x₂ = vt + L in the above expression.

$$\begin{gathered} c\left(t-t_2\right)=v{t}_2+L \\ t-\frac{L}{c}=t_2\left(1+\frac{v}{c}\right) \\ {t}_2=\frac{t-{\displaystyle \frac{L}{c}}}{1+\frac{v}{c}} \end{gathered}$$

Calculate the corresponding retarded position x₁ .

$\begin{array}{rcl}{x}_1& =& v{t}_1\\ x_1& =& v\left(\frac{t}{1+{\displaystyle \frac{v}{c}}}\right)\\ x_1& =& \frac{vt}{1+v/c}\end{array}$

Calculate the corresponding retarded position x₂ .

$$\begin{gathered} x_2=v\left(\frac{t-{\displaystyle \frac{L}{c}}}{1+{\displaystyle \frac{v}{c}}}\right)+L \\ {x}_2=\frac{vt-vL/c+L+vL/c}{1+v/c} \\ {x}_2=\frac{vt+L}{1+v/c} \end{gathered}$$

$V=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{vt}\right)$Calculate the retarded potential.

$$\begin{gathered} V\left(0,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}{\int }_{x_1}^{x_2}\frac{\lambda }{x}dx \\ V\left(0,t\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}{\int }_{x_1}^{x_2}\frac{1}{x}dx \\ V\left(0,t\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}in\left(\frac{x_2}{x_1}\right) \\ V\left(0,t\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}in\left(\frac{vt+L}{vtt}\right) \end{gathered}$$

If L<< vt , re-write the above expression.

$$\begin{gathered} V\left(0,t\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}in\left(1+\frac{L}{vt}\right) \\ V\left(0,t\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\frac{L}{vt} \\ V\left(0,t\right)=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{vt}\right) \end{gathered}$$

Write the expression for the Lienard-Wiechart potential.

role="math" localid="1657773654694" $V=\frac{q}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{r}-\mathrm{r}·{\displaystyle \frac{\mathrm{v}}{\mathrm{c}}}\right)}$ …… (1)

Here, it is known that:

$$\begin{gathered} \mathrm{r}-\mathrm{r}·\frac{\mathrm{v}}{\mathrm{c}}=v{t}_r+\frac{{v^2}_{tr}}{c} \\ \mathrm{r}-\mathrm{r}·\frac{\mathrm{v}}{\mathrm{c}}v\left(1+\frac{v}{c}\right)t_r \\ \mathrm{r}-\mathrm{r}·\frac{\mathrm{v}}{\mathrm{c}}=v\left(1+\frac{v}{c}\right)\frac{t}{1+\frac{v}{c}} \\ \mathrm{r}-\mathrm{r}·\frac{\mathrm{v}}{c}=vt \end{gathered}$$

Substitute $r-r·\frac{v}{c}=vt$in equation (1).

$$\begin{gathered} V=\frac{q}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{vt}\right)} \\ V=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{vt}\right) \end{gathered}$$

Therefore, the retarded scalar potential at the origin as a function of time is $V=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{vt}\right)$.