Charge Density inside Conductor

[th5418]

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 \rho_f will dissolve exponentially with time. Express the characteristic time needed to dissolve the charge in terms of the conductor's dielectric constant \epsilon and the conductivity \sigma.

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

3. The attempt at a solution
I know what the solution should be..
\rho (t) = \rho_0 e^{t}
where t=\frac{\epsilon}{\sigma}

[gabbagabbahey]

I think I should use charge conservation. I'm not sure...
delJ + \frac{d\rho}{dt} = 0

Assuming you mean \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} and \sigma that might help you here?:wink:


I know what the solution should be..
\rho (t) = \rho_0 e^{t}
where t=\frac{\epsilon}{\sigma}


Surely you mean \rho(t)=\rho_0 e^{-t/\tau}, where \tau\equiv\epsilon/\sigma...right?

[Reshma]

Just a hint: Rearranging the equation
\vec \nabla \cdot \vec J = -\frac{d \rho}{dt}

Can you express \vec J in terms of \rho (t) ?

[caduceus]

\vec \nabla \cdot \vec J = -\frac{\partial\rho}{\partial t}

and you should use the relation:

\vec J = \sigma\vec E

where \vec\nabla \cdot \vec E=\frac{\rho}{\epsilon}

[Count Iblis]

Ohm's law:

\vec{J}=\sigma \vec{E}

is not valid on the relevant time scale for this problem.

[caduceus]

No! Ohm's law is still valid. Only when the time is shorter than \tau (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time \tau, electrostatic equilibrium is reached, and finding \tau is our concern.

[Count Iblis]

No! Ohm's law is still valid. Only when the time is shorter than \tau (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time \tau, electrostatic equilibrium is reached, and finding \tau is our concern.

Ohm's law is valid on time scales much longer than the typical collision time. The time scale \tau in this problem will be many orders of magnitude less than that.