A uniformly charged rod (length L, charge density λ) slides out the x axis at constant speed υ. At time t = 0 the back end passes the origin (so its position as a function of time is x = υt, while the front end is at x = vt + L). Find the retarded scalar potential at the origin, as a function o

Question

A uniformly charged rod (length L, charge density λ) slides out the x axis at constant speed υ . At time t = 0 the back end passes the origin (so its position as a function of time is x = υt, while the front end is at x = υt + L). Find the retarded scalar potential at the origin, as a function of time, for t > 0. [First determine the retarded time [latex]t_{1}[/latex] for the back end, the retarded time [latex]t_{2}[/latex] for the front end, and the corresponding retarded positions [latex] x_{1} ext { and } x_{2}[/latex] .] Is your answer consistent with the Liénard-Wiechert potential, in the point charge limit [latex](L \ll v t, ext { with } \lambda L=q)[/latex]? Do not assume [latex]v \ll c[/latex] .

Accepted Answer

[latex]c\left(t-t_{1} ight)=v t_{1}, t=t_{1}(1+v / c), t_{1}=\frac{t}{1+v / c} . \quad c\left(t-t_{2} ight)=v t_{2}+L, t-L / c=t_{2}(1+v / c), \quad t_{2}=\frac{t-L / c}{1+v / c}[/latex]

[latex]x_{1}=v t_{1}=\frac{v t}{1+v / c} \cdot \quad x_{2}=v t_{2}+L=\frac{v(t-L / c)}{1+v / c}+L=\frac{v t-v L / c+L+v L / c}{1+v / c}=\frac{v t+L}{1+v / c}[/latex] .

[latex]V( 0 , t)=\frac{1}{4 \pi \epsilon_{0}} \int_{x_{1}}^{x_{2}} \frac{\lambda}{x} d x=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left(\frac{x_{2}}{x_{1}} ight)=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left(\frac{v t+L}{v t} ight)[/latex] .

[latex] ext { If } L \ll v t, V=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left(1+\frac{L}{v t} ight) \approx \frac{\lambda}{4 \pi \epsilon_{0}} \frac{L}{v t}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{v t} ight)[/latex] . The [latex] ext { Liénard }[/latex]-Wiechert potential is

[latex]V=\frac{q}{4 \pi \epsilon_{0}(ᴫ- ᴫ \cdot v / c)}[/latex] . Here

[latex](ᴫ- ᴫ \cdot v / c)=v t_{r}+v^{2} t_{r} / c=v(1+v / c) t_{r}=v(1+v / c) \frac{t}{1+v / c}=v t[/latex] .

[latex] ext { (In the limit } \left.L ightarrow 0, t_{r}=t_{1}=t_{2} . ight) \quad ext { So } V=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{v t} ight)[/latex].

Question 10.30: A uniformly charged rod (length L, charge density λ) slides ...

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