# Charge distribution and potential in a 1-dimensional quasistatic syste

1. Nov 12, 2013

### christianpoved

1. The problem statement, all variables and given/known data

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 $\frac{d}{dt}=0$.

2. Relevant 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)

3. 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!
$$\frac{df}{f^{1/4}}=\sqrt{-4k}dx$$
$$\frac{4}{3} f^{3/4}=\sqrt{-4k}x +C$$
$$f={\frac{3}{4}[\sqrt{-4k}x +C]}^{4/3}$$

Last edited: Nov 12, 2013
2. Nov 13, 2013

### Staff: Mentor

Both v and ϕ depend on x here. And I would expect that m, if necessary at all, depends on x as well.

I am not sure if you are supposed to consider a time-dependent ρ apart from the uniform motion, I think you do not have to. Just use ρ(x) and v(x) as unknown but constant functions. Then there is no motion where you would have to consider conservation laws.

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