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!

Homework Help: 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


    User Avatar
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook