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christianpoved
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Homework Statement
suppose you have an 1-dimensional system with a charge distribution ##\rho(x)## (not given) moving with an speed ##v(x)##, calculate the potential ##\phi(x)## and the charge distribution ##\rho(x)## in the quasistatic limit ##\frac{d}{dt}=0##.
Homework Equations
##\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)
The Attempt at a Solution
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}## 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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