Please explain the concept and the steps marked in green.Please explain in detail.Example $10\mathrm{.}1$

Find the charge and current distributions that would give rise to the potentials

$V=0,=\{\begin{array}{ccc}\frac{{}_{0}k}{4c}(ct-|x|{)}^2\text{h}\text{a}\text{t}& (z),\text{f}\text{o}\text{r}& |x|ct,\end{array}$

where $k$ is a constant, and $c=\frac{1}{\sqrt[\phantom{2}]{lo{n}_0{}_0}}.$

Solution: First we'll determine the electric and magnetic fields, using Eqs. $10\mathrm{.}2$ and $10\mathrm{.}3$:

$E=-\frac{\text{d}\text{e}\text{l}\text{A}}{\text{d}\text{e}\text{l}\text{t}}=-\frac{{}_{0}k}{2}(ct-|x|)hat(z),$

$B=$grad$A=-\frac{{}_{0}k}{4c}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}(ct-|x|{)}^2$hat$(y)=+-\frac{{}_{0}k}{2c}(ct-|x|)hat(y),$

$($plus for $x>0,$ minus for These are for $|x|>ct,E=B=0$grad$*E=0$;$,$grad$*B=0$;$,$grad$E=\frac{{}_{0}k}{2}$hat$(y)$;$,$grad$B=-\frac{{}_{0}k}{2c}$hat$(z)$;

$\frac{\text{d}\text{e}\text{l}\text{E}}{\text{d}\text{e}\text{l}\text{t}}=-\frac{{}_{0}kc}{2}$hat$(z)$;$,\frac{\text{d}\text{e}\text{l}\text{B}}{\text{d}\text{e}\text{l}\text{t}}=+-\frac{{}_{0}k}{2}$hat$(y).JBx=0$Kyzkthat$(y)=K$hat$(x),K=$kthat$(z).zx=0t=0t|x|>ct|x|$; when $|x|>ct,E=B=0$

$(Fig\mathrm{.}10\mathrm{.}1).$ Calculating every derivative $in$ sight, I find

grad$*E=0$;$,$grad$*B=0$;$,$grad$E=\frac{{}_{0}k}{2}$hat$(y)$;$,$grad$B=-\frac{{}_{0}k}{2c}$hat$(z)$;

$\frac{\text{d}\text{e}\text{l}\text{E}}{\text{d}\text{e}\text{l}\text{t}}=-\frac{{}_{0}kc}{2}$hat$(z)$;$,\frac{\text{d}\text{e}\text{l}\text{B}}{\text{d}\text{e}\text{l}\text{t}}=+-\frac{{}_{0}k}{2}$hat$(y).$

$As$ you can easily check, Maxwell's equations are all satisfied, with $$ and $J$ both zero. Notice,

however, that $B$ has a discontinuity $atx=0,$ and this signals the presence $of$ a surface current

$Kin$ the $yz$ plane; boundary condition $(iv)inEq.7\mathrm{.}63$ gives

kthat$(y)=K$hat$(x),$

and hence

$K=$kthat$(z).$

Evidently $we$ have here a uniform surface current flowing $in$ the $z$ direction over the plane

$x=0,$ which starts $upatt=0,$ and increases $in$ proportion $tot.$ Notice that the news travels

out $(in$ both directions$)at$ the speed $of$ light: for points $|x|>ct$ the message $(that$ current $is$

now flowing$)$ has not yet arrived, $so$ the fields are zero.