Suppose you have an 1-dimensional system with a charge distribution $\rho(x)$ (not given) moving with an speed $v(x)$ (not given), calculate the potential $\phi(x)$ and the charge distribution $\rho(x)$ in the quasistatic limit $\frac{d}{dt}=0$.


$\frac{d^{2} \phi}{dx^{2}}=-\rho/ \varepsilon_{0}$ (Poisson equation)

$j=\rho v$

$\frac{d}{dx}(\rho v)=0$ (Continuity equation)

$\frac{1}{2} mv^{2}=q\phi$(Energy Conservation)

My attempt for solving it was: From the energy conservation equation we get that $$\frac{1}{v}=\sqrt{\frac{m}{2q}} \phi^{-1/2}$$ Continuity equation tells us that $j$ is constant, then $$ \rho = \frac{j}{v}=j\sqrt{\frac{m}{2q}} \phi^{-1/2}$$ Using Laplace equation $$-\varepsilon_{0}\frac{d^{2} \phi}{dx^{2}}=j\sqrt{\frac{m}{2q}} \phi^{-1/2}$$ Then $$\frac{d^{2} \phi}{dx^{2}}+\frac{j}{\varepsilon_{0}}\sqrt{\frac{m}{2q}} \phi^{-1/2}=0$$ This is just an equation of the form $$ f^{\prime\prime}+kf^{-1/2}=0$$ Multiplying by $f^{\prime}$ and integrating $$ \int f^{\prime}f^{\prime\prime}dx+k\int f^{\prime}f^{-1/2}dx=0$$ $$ \frac{1}{2}(f^{\prime})^2+2k\sqrt{f}=0$$ $$ (f^{\prime})^2=-4k\sqrt{f}$$ $$ f^{\prime}=\sqrt{-4k} f^{1/4}$$ And here is my problem, i have the $\sqrt{-4k}$ term that in general is not real! $$ \frac{df}{f^{1/4}}=\sqrt{-4k}dx $$ $$ \frac{4}{3} f^{3/4}=\sqrt{-4k}x +C $$ $$ f={\frac{3}{4}[\sqrt{-4k}x +C]}^{4/3} $$


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