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Charge Density inside Conductor

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    We know that free charges inside a conductor will eventually move to the conductor surface. Consider a free charge initially placed inside a conductor at t=0. Show that the free charge density [tex]\rho_f[/tex] will dissolve exponentially with time. Express the characteristic time needed to dissolve the charge in terms of the conductor's dielectric constant [tex]\epsilon[/tex] and the conductivity [tex]\sigma[/tex].

    2. Relevant equations
    I think I should use charge conservation. I'm not sure...
    [tex]delJ + \frac{d\rho}{dt} = 0[/tex]

    3. The attempt at a solution
    I know what the solution should be..
    [tex]\rho (t) = \rho_0 e^{t} [/tex]
    where [tex]t=\frac{\epsilon}{\sigma}[/tex]
     
  2. jcsd
  3. Dec 3, 2009 #2

    gabbagabbahey

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    Assuming you mean [itex]\mathbf{\nabla}\cdot\textbf{J}+\frac{d\rho}{dt}=0[/tex] (i.e. divJ not "delJ" ), that seems like a good start to me....is there some relationship between [itex]\textbf{J}[/itex] and [itex]\sigma[/itex] that might help you here?:wink:


    Surely you mean [itex]\rho(t)=\rho_0 e^{-t/\tau}[/itex], where [itex]\tau\equiv\epsilon/\sigma[/itex]...right?
     
  4. Dec 3, 2009 #3
    Just a hint: Rearranging the equation
    [tex]\vec \nabla \cdot \vec J = -\frac{d \rho}{dt}[/tex]

    Can you express [tex] \vec J[/tex] in terms of [itex]\rho (t)[/itex] ?
     
  5. Dec 30, 2009 #4
    [tex]\vec \nabla \cdot \vec J = -\frac{\partial\rho}{\partial t}[/tex]

    and you should use the relation:

    [tex]\vec J = \sigma\vec E[/tex]

    where [tex]\vec\nabla \cdot \vec E=\frac{\rho}{\epsilon}[/tex]
     
  6. Dec 30, 2009 #5
    Ohm's law:

    [tex]\vec{J}=\sigma \vec{E}[/tex]

    is not valid on the relevant time scale for this problem.
     
  7. Dec 31, 2009 #6
    No! Ohm's law is still valid. Only when the time is shorter than [tex]\tau[/tex] (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time [tex]\tau[/tex], electrostatic equilibrium is reached, and finding [tex]\tau[/tex] is our concern.
     
    Last edited: Dec 31, 2009
  8. Dec 31, 2009 #7
    Ohm's law is valid on time scales much longer than the typical collision time. The time scale [tex]\tau[/tex] in this problem will be many orders of magnitude less than that.
     
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