1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charge distribution and potential in a 1-dimensional quasistatic syste

  1. Nov 12, 2013 #1
    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. jcsd
  3. Nov 13, 2013 #2

    mfb

    User Avatar
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Charge distribution and potential in a 1-dimensional quasistatic syste
Loading...